Integration techniques

<1>1. \(\int (\ln x)^2 \, dx = x (\ln x)^2 - 2x \ln x + 2x + C\).

<1>2. For \(b \neq 0\), the reduction formula \(\int x^a e^{bx} \, dx = \frac{1}{b} x^a e^{bx} - \frac{a}{b} \int x^{a-1} e^{bx} \, dx\) holds.

<1>3. The following cases of the reduction formula hold.

<1>1. \(\int \frac{\sec^3(x) + e^{\sin(x)}}{\sec(x)} \, dx = \tan(x) + e^{\sin(x)} + C\).

<1>2. \(\int \frac{1}{\tan(x) + \cot(x)} \, dx = \frac{1}{2}\sin^2(x) + C = -\frac{1}{2}\cos^2(x) + C'\).

Take \(|x| < 5\).

<1>1. \(\int \frac{x^3}{\sqrt{25-x^2}} \, dx = \frac{1}{3}(25-x^2)^{3/2} - 25\sqrt{25-x^2} + C = \frac{1}{3}\sqrt{25-x^2}(-x^2 - 50) + C\).

<1>2. The trigonometric substitution \(x = 5\sin(\theta)\) gives the same antiderivative.

Take \(x \le 4\).

<1>1. Make the substitution \(u = 4 - x\). Then \(du = -dx \implies dx = -du\).

<1>2. Transform and evaluate the integral: \[\int \sqrt{4-x} \, dx = \int u^{1/2} (-du) = -\int u^{1/2} \, du = -\frac{2}{3} u^{3/2} + C.\]

<1>3. Substitute back \(u = 4-x\): \[\int \sqrt{4-x} \, dx = -\frac{2}{3} (4-x)^{3/2} + C.\]

<1>1. Make the substitution \(u = x^2 + 9\). Then \(du = 2x \, dx \implies x \, dx = \frac{1}{2} \, du\).

<1>2. Rewrite and evaluate the integral: \[\int \frac{x}{x^2+9} \, dx = \int \frac{1}{u} \left(\frac{1}{2} \, du\right) = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C.\]

<1>3. Substitute back \(u = x^2 + 9\): Since \(x^2 + 9 > 0\) for all \(x \in \mathbb{R}\), \(|u| = x^2 + 9\), so: \[\int \frac{x}{x^2+9} \, dx = \frac{1}{2} \ln(x^2+9) + C.\]

Take \(|x| < 2\).

<1>1. Split the integral into two parts: \[\int \frac{x+1}{\sqrt{4-x^2}} \, dx = \int \frac{x}{\sqrt{4-x^2}} \, dx + \int \frac{1}{\sqrt{4-x^2}} \, dx.\]

<1>2. \(\int \frac{x}{\sqrt{4-x^2}} \, dx = -\sqrt{4-x^2} + C_1\).

<1>3. \(\int \frac{1}{\sqrt{4-x^2}} \, dx = \arcsin\left(\frac{x}{2}\right) + C_2\).

<1>4. Combining results: \[\int \frac{x+1}{\sqrt{4-x^2}} \, dx = -\sqrt{4-x^2} + \arcsin\left(\frac{x}{2}\right) + C.\]

<1>1. With \(u = \sin^2(x)\), \(du = \sin(2x) \, dx\).

<1>2. Transform and evaluate the integral in terms of \(u\): \[\int e^{\sin^2(x)} \sin(2x) \, dx = \int e^u \, du = e^u + C.\]

<1>3. Substitute back \(u = \sin^2(x)\): \[\int e^{\sin^2(x)} \sin(2x) \, dx = e^{\sin^2(x)} + C.\]

Take \(a > 0\) and \(x > a\).

<1>1. Substitute \(x = a \sec(\theta)\) with \(\theta \in (0, \pi/2)\).

<1>2. Transform and integrate: \[\int \frac{\sqrt{x^2-a^2}}{x} \, dx = \int \frac{a \tan(\theta)}{a \sec(\theta)} \cdot a \sec(\theta)\tan(\theta) \, d\theta = a \int \tan^2(\theta) \, d\theta.\]

<1>3. Evaluate \(a \int \tan^2(\theta) \, d\theta = a (\tan(\theta) - \theta) + C\).

<1>4. Express the antiderivative in terms of \(x\): \[\int \frac{\sqrt{x^2-a^2}}{x} \, dx = \sqrt{x^2-a^2} - a \operatorname{arcsec}\left(\frac{x}{a}\right) + C = \sqrt{x^2-a^2} - a \arctan\left(\frac{\sqrt{x^2-a^2}}{a}\right) + C.\]

<1>1. The substitution \(u = \sin(x)\) gives \(\int \frac{\cos(x)}{\sin^2(x)} \, dx = -\csc(x) + C\).

<1>2. The identity \(\frac{\cos(x)}{\sin^2(x)} = \cot(x)\csc(x)\) gives the same antiderivative.

<1>3. Q.E.D.

<1>1. For \(x \in [\pi/3, \pi/2]\), \(\frac{1-\cos(x)}{\sin(x)} = \tan\left(\frac{x}{2}\right) = \csc(x) - \cot(x)\).

<1>2. \(\int \frac{1-\cos(x)}{\sin(x)} \, dx = -2 \ln|\cos(x/2)| + C = -\ln(1+\cos(x)) + C\).

<1>3. \[\int_{\pi/3}^{\pi/2} \frac{1-\cos(x)}{\sin(x)} \, dx = \left[ -\ln(1+\cos(x)) \right]_{\pi/3}^{\pi/2} = \boxed{\ln\left(\frac{3}{2}\right)}.\]

<1>1. \(\int \frac{x^3 + 4x^2}{x^2 + 4x + 3} \, dx = \frac{1}{2}x^2 + \frac{3}{2}\ln|x+1| - \frac{9}{2}\ln|x+3| + C\).

<1>2. \(\int \frac{2x^3 + 2x^2 - 9x - 1}{x^2 + x - 6} \, dx = x^2 + 2\ln|x+3| + \ln|x-2| + C\).

<1>3. \(\int \frac{3x^3 - 12x^2 + 15x - 5}{(x-2)^2} \, dx = \frac{3}{2}x^2 + 3\ln|x-2| - \frac{1}{x-2} + C\).

<1>1. The antiderivative of \(\frac{1}{x\sqrt{x^2-1}}\) for \(x > 1\) is \(\operatorname{arcsec}(x) + C = \arccos(1/x) + C\).

<1>2. The definite integral is an improper integral at the lower limit \(x=1\), defined by \(\lim_{t \to 1^+} \int_t^2 \frac{1}{x\sqrt{x^2-1}} \, dx\).

<1>3. \[\int_1^2 \frac{1}{x\sqrt{x^2-1}} \, dx = \lim_{t \to 1^+} [\operatorname{arcsec}(x)]_t^2 = \operatorname{arcsec}(2) - \lim_{t \to 1^+} \operatorname{arcsec}(t) = \frac{\pi}{3} - 0 = \boxed{\frac{\pi}{3}}.\]

<1>1. For all \(x \in [0, \pi]\), \(1 + \cos(2x) = 2\cos^2(x)\), hence \(\sqrt{1 + \cos(2x)} = \sqrt{2}|\cos(x)|\).

<1>2. Partitioning the integral over intervals of constant sign of \(\cos(x)\): \[\int_0^\pi |\cos(x)| \, dx = \int_0^{\pi/2} \cos(x) \, dx + \int_{\pi/2}^\pi (-\cos(x)) \, dx.\]

<1>3. Computing each piece:

<1>4. Conclusion: \[\int_0^\pi \sqrt{1+\cos(2x)} \, dx = \sqrt{2} \int_0^\pi |\cos(x)| \, dx = 2\sqrt{2}.\]

<1>1. \(\int \frac{x e^x \ln x - e^x}{x (\ln x)^2} \, dx = \frac{e^x}{\ln x} + C\).

<1>2. \(\int (\tan x + \cot x)^2 \, dx = \tan(x) - \cot(x) + C\).

<1>1. \(\int_0^1 \frac{1}{\sqrt{x(1-x)}} \, dx = \pi\) and \(\int_{e^{1/2}}^{e^{3/4}} \frac{1}{x\sqrt{\ln x(1-\ln x)}} \, dx = \frac{\pi}{6}\).

<1>2. \(\int \frac{x}{1-x^2+\sqrt{1-x^2}} \, dx = -\ln(\sqrt{1-x^2}+1) + C\).

<1>3. \(\int_1^\infty \frac{1}{x(x^2+1)} \, dx = \frac{1}{2}\ln(2)\).

<1>1. Use product-to-sum.

<1>2. Use product-to-sum again.

<1>3. Reduce the triple product.

Let \(u=\cos x\), so \(du=-\sin x\,dx\). Then \[ \int \sin x\,\cos(\cos x)\,dx =-\int\cos u\,du =\boxed{-\sin(\cos x)+C}. \]

For a continuous function \(\Phi\), the substitution \(u=1/x\), \(du=-x^{-2}\,dx\) gives \[ \int \frac{\Phi(1/x)}{x^2}\,dx=-\int \Phi(u)\,du. \]

For the first integral, integration by parts with \(u=\arcsin(1/x)\) and \(dv=x\,dx\) gives \[ \boxed{\int x\arcsin\!\left(\frac1x\right)\,dx =\frac12\left(x^2\arcsin\!\left(\frac1x\right)+\sqrt{x^2-1}\right)+C.} \]

For the second, \[ \boxed{\int x\arctan x\,dx =\frac12\left((x^2+1)\arctan x-x\right)+C.} \] Therefore \[ \boxed{\int_0^1x\arctan x\,dx=\frac\pi4-\frac12.} \]

Finally, integration by parts with \(u=x\) and \(dv=\sin(2x)dx\) gives \[ \boxed{\int x\sin(2x)\,dx =-\frac12x\cos(2x)+\frac14\sin(2x)+C.} \]

<1>1. Evaluate the infinite nested radical.

<1>2. Evaluate the logarithmic sine integral.

<1>3. Evaluate the squared-sinc integral.

<1>4. Pass the limit through the integral.

<1>5. Evaluate the exponentially damped sine integral.

<1>6. Evaluate the Cauchy-kernel cosine transform.

<1>7. Integrate the square-root quotient.

Let \(u=e^x+1\), so \(du=e^x\,dx\). Then \[ \boxed{\int \frac{e^x}{e^x+1}\,dx=\ln(e^x+1)+C.} \]

Use the tangent half-angle substitution \[ u=\tan\frac x2, \qquad \sin x=\frac{2u}{1+u^2}, \qquad \cos x=\frac{1-u^2}{1+u^2}, \qquad dx=\frac{2\,du}{1+u^2}. \]

<1>1. The minus-sign numerator.

<1>2. The plus-sign numerator.

<1>3. The denominator \(1+\sin x+\cos x\).

Standard reduction and \(u\)-substitutions give \[ \boxed{\int\sec^3x\,dx =\frac12\bigl(\sec x\tan x+\ln|\sec x+\tan x|\bigr)+C,} \] \[ \boxed{\int\sec^4x\,dx=\tan x+\frac13\tan^3x+C,} \] \[ \boxed{\int\tan^2x\,dx=\tan x-x+C,} \] \[ \boxed{\int\tan^3x\,dx=\frac12\tan^2x+\ln|\cos x|+C,} \] \[ \boxed{\int\tan^4x\,dx=\frac13\tan^3x-\tan x+x+C.} \] For the mixed powers, use \(u=\sec x\) in the first and \(u=\tan x\) in the second: \[ \boxed{\int\sec^3x\tan^3x\,dx=\frac15\sec^5x-\frac13\sec^3x+C,} \] \[ \boxed{\int(\tan^4x+\tan^2x)dx=\frac13\tan^3x+C.} \] Finally, \[ \frac{2\sin^2x}{\cos^3x}=2(\sec^3x-\sec x), \] so \[ \boxed{\int\frac{2\sin^2x}{\cos^3x}dx =\sec x\tan x-\ln|\sec x+\tan x|+C.} \] Using \((\sec x-\tan x)(\sec x+\tan x)=1\), this is equivalently \[ \sec x\tan x+\ln|\sec x-\tan x|+C. \]

Using elementary partial fractions, \[ \frac{x}{x-1}=1+\frac1{x-1}, \] so \[ \boxed{x+\ln|x-1|+C.} \] Also \[ \frac{7x+5}{(x+2)(x-1)}=\frac3{x+2}+\frac4{x-1}, \] hence \[ \boxed{3\ln|x+2|+4\ln|x-1|+C.} \] Next, \[ \frac{x}{(x-3)(x+2)}=\frac{3/5}{x-3}+\frac{2/5}{x+2}, \] so \[ \boxed{\frac35\ln|x-3|+\frac25\ln|x+2|+C.} \] Further, \[ \frac{1+x^2}{1-x^2}=-1+\frac1{x+1}-\frac1{x-1}, \] therefore \[ \boxed{-x+\ln|x+1|-\ln|x-1|+C.} \] For the repeated factor, \[ \frac1{x(x-1)^2}=\frac1x-\frac1{x-1}+\frac1{(x-1)^2}, \] which gives \[ \boxed{\ln|x|-\ln|x-1|-\frac1{x-1}+C.} \] Finally, \[ \frac{x}{(x+5)^2}=\frac1{x+5}-\frac5{(x+5)^2}, \] so \[ \boxed{\ln|x+5|+\frac5{x+5}+C.} \]

With \(u=x^{1/a}\), so \(x=u^a\) and \(dx=a u^{a-1}du\), \[ \int e^{x^{1/a}}\,dx=a\int u^{a-1}e^u\,du. \] For positive integer \(a\) this is elementary by repeated integration by parts.

For \(a=2\), \[ \boxed{\int e^{\sqrt x}\,dx=2(\sqrt x-1)e^{\sqrt x}+C.} \] For \(a=3\), \[ \boxed{\int e^{\sqrt[3]x}\,dx=3\left(x^{2/3}-2x^{1/3}+2\right)e^{x^{1/3}}+C.} \]

Integration by parts twice gives \[ \boxed{\int e^x\sin x\,dx=\frac12e^x(\sin x-\cos x)+C,} \] \[ \boxed{\int e^x\cos x\,dx=\frac12e^x(\cos x+\sin x)+C.} \] Finally, putting \(u=\sqrt x\) gives \[ \int \frac{\sin(\sqrt x)e^{\sqrt x}}{\sqrt x}\,dx =2\int e^u\sin u\,du =\boxed{e^{\sqrt x}(\sin\sqrt x-\cos\sqrt x)+C.} \]

The standard Gaussian integral is \[ \int_{-\infty}^{\infty}e^{-u^2}\,du=\sqrt\pi. \] For the shifted-scaled Gaussian, put \(u=(x-\mu)/(\sqrt2\,|\sigma|)\): \[ \boxed{\int_{-\infty}^{\infty}e^{-(x-\mu)^2/(2\sigma^2)}\,dx=\sqrt{2\pi}\,|\sigma|.} \] Thus \[ \boxed{\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt\pi}, \qquad \boxed{\int_{-\infty}^{\infty}e^{-x^2/2}\,dx=\sqrt{2\pi}}. \] Finally, \[ -x^2+6x-3=-(x-3)^2+6, \] so \[ \boxed{\int_{-\infty}^{\infty}e^{-x^2+6x-3}\,dx=e^6\sqrt\pi.} \]

For the first integral, let \(u=\ln x\), so \(du=dx/x\): \[ \int\frac{dx}{x(\ln x)^2} =\int u^{-2}\,du =\boxed{-\frac1{\ln x}+C}. \] Thus \[ \int_2^\infty\frac{dx}{x(\ln x)^2} =\lim_{R\to\infty}\left[-\frac1{\ln x}\right]_{2}^{R} =\boxed{\frac1{\ln2}}. \] For the second antiderivative, let \(u=1/x\), so \(du=-dx/x^2\): \[ \boxed{\int \frac1{x^2}\cos\!\left(\frac1x\right)\,dx=-\sin\!\left(\frac1x\right)+C.} \]

Since \[ \frac{d}{dx}2018^x=(\ln 2018)2018^x, \] we have \[ \boxed{\int 2018^x\,dx=\frac{2018^x}{\ln 2018}+C.} \]

Let \(u=x^2\), so \(du=2x\,dx\). Then \[ \boxed{\int\frac{x}{1+x^4}\,dx=\frac12\arctan(x^2)+C.} \]

Using \(\sin x/\cos^2x=\tan x\sec x\) and \((\sec x)'=\sec x\tan x\), \[ \boxed{\int\frac{\sin x}{\cos^2x}\,dx=\sec x+C.} \] Also \((\csc(ax))'=-a\csc(ax)\cot(ax)\), so \[ \boxed{\int\csc(ax)\cot(ax)\,dx=-\frac1a\csc(ax)+C.} \]

Since \[ \frac1{e^x+e^{-x}}=\frac{e^x}{1+e^{2x}}, \] letting \(u=e^x\) gives \[ \boxed{\int\frac{dx}{e^x+e^{-x}}=\int\frac{e^x}{1+e^{2x}}\,dx=\arctan(e^x)+C.} \] For the second type, let \(u=e^{-x}\). Since \(du=-e^{-x}dx\) and \[ \sqrt{e^{2x}-1}=\frac{\sqrt{1-u^2}}{u} \qquad (u>0), \] we get \[ \int\frac{dx}{\sqrt{e^{2x}-1}} =-\int\frac{du}{\sqrt{1-u^2}} =\boxed{-\arcsin(e^{-x})+C}. \] Equivalently, with \(v=\sqrt{e^{2x}-1}\), \[ \boxed{\int\frac{dx}{\sqrt{e^{2x}-1}}=\arctan\sqrt{e^{2x}-1}+C'}, \] and the two primitives differ by the constant \(\pi/2\) on \(x>0\).

<1>1. Integrate \(1/(\sin x+\cos x)\).

<1>2. Integrate \(\sin x/(1+\sin x)\).

<1>3. Integrate \(\sin x/(1-\sin x)\).

Integrating the first integral by parts twice yields \[ \boxed{\int e^x\cos x\,dx=\frac12e^x(\cos x+\sin x)+C.} \] For the second, let \(u=\ln x\), so \(dx=e^u du\). Then \[ \int\sin(\ln x)\,dx=\int e^u\sin u\,du =\boxed{\frac x2\left(\sin(\ln x)-\cos(\ln x)\right)+C.} \]

Since \(\ln\sqrt{x^2+1}=\tfrac12\ln(x^2+1)\), integration by parts gives \[ \begin{aligned} \int \ln\sqrt{x^2+1}\,dx &=\frac x2\ln(x^2+1)-\int\frac{x^2}{x^2+1}\,dx\\ &=\boxed{\frac x2\ln(x^2+1)-x+\arctan x+C.} \end{aligned} \]

For the second integral, integrate by parts with \[ u=\arctan(e^x),\qquad dv=2e^{-2x}\,dx, \] so \(du=e^x(1+e^{2x})^{-1}dx\) and \(v=-e^{-2x}\). Thus \[ I=-e^{-2x}\arctan(e^x)+\int\frac{e^{-x}}{1+e^{2x}}\,dx. \] Using \[ \frac{e^{-x}}{1+e^{2x}}=e^{-x}-\frac{e^x}{1+e^{2x}}, \] we obtain \[ \boxed{I=-e^{-2x}\arctan(e^x)-e^{-x}-\arctan(e^x)+C.} \]

<1>1. On the domain \(\{x \in \mathbb{R} : \sin(x) \neq 0, \cos(x) \neq 0\}\) of the integrand, \(\frac{\sin^3(x)}{\cos(x) - \cos^3(x)} = \tan(x)\).

<1>2. \(\int \tan(x) \, dx = -\ln|\cos(x)| + C\).

<1>3. Q.E.D.

For the first integral, split \[ \frac{x^2+2x+1}{x\sqrt{x^2-1}} =\frac{x}{\sqrt{x^2-1}}+\frac2{\sqrt{x^2-1}}+\frac1{x\sqrt{x^2-1}}. \] The three primitives are respectively \[ \sqrt{x^2-1},\qquad 2\ln|x+\sqrt{x^2-1}|, \qquad \arctan\sqrt{x^2-1}. \] Thus \[ \boxed{\sqrt{x^2-1}+2\ln|x+\sqrt{x^2-1}|+\arctan\sqrt{x^2-1}+C.} \]

For the second integral, write \[ x^2-4x+8=(x-2)^2+4, \qquad x+16=(x-2)+18. \] Then \[ \boxed{\int\frac{x+16}{\sqrt{x^2-4x+8}}\,dx =\sqrt{x^2-4x+8}+18\ln\!\left|x-2+\sqrt{x^2-4x+8}\right|+C.} \]

The corresponding primitives are, respectively, \[ \boxed{x-\arctan x+C,} \] \[ \boxed{4x+\frac12\ln(x^2+1)+3\arctan x+C,} \] \[ \boxed{3\ln|x|+\arctan x+C,} \] \[ \boxed{-3\ln|x|+\frac32\ln(x^2+1)+2\arctan x+C,} \] \[ \boxed{\ln|x-5|+\frac32\ln(x^2+1)+4\arctan x+C,} \] \[ \boxed{3\ln|x-5|+2\arctan x+C,} \] \[ \boxed{-2\ln|x-1|-\frac1{x-1}+\ln(x^2+1)+\arctan x+C,} \] \[ \boxed{\frac12x^2+x+3\ln|x+1|+2\arctan x+C,} \] and \[ \boxed{x^2+\ln(x^4+1)+\arctan(x^2)+C.} \] Each follows from the partial-fraction decomposition of its integrand; differentiating each displayed primitive recovers the corresponding rational integrand.

For the antiderivative, set \(x=\sin u\). Then \(dx=\cos u\,du\) and \[ \sqrt{1-x^2}=\cos u \] on the principal substitution interval, so \[ \int\sqrt{1-x^2}\,dx =\int\cos^2u\,du =\frac12\left(u+\sin u\cos u\right)+C. \] Thus \[ \boxed{\int\sqrt{1-x^2}\,dx=\frac12\left(\arcsin x+x\sqrt{1-x^2}\right)+C.} \]

The graph of \(y=\sqrt{16-x^2}\) is the upper semicircle of radius \(4\), so \[ \boxed{\int_{-4}^{4}\sqrt{16-x^2}\,dx=8\pi.} \] Finally, \[ \int_{-1}^{1}(1+x)\sqrt{1-x^2}\,dx =\int_{-1}^{1}\sqrt{1-x^2}\,dx +\int_{-1}^{1}x\sqrt{1-x^2}\,dx. \] The second integrand is odd, hence its integral is \(0\), while the first integral is the area of a semicircle of radius \(1\). Therefore \[ \boxed{\int_{-1}^{1}(1+x)\sqrt{1-x^2}\,dx=\frac\pi2.} \]

For the first, put \(u=\sqrt{1+\ln x}\). Then \(\ln x=u^2-1\) and \(dx/x=2u\,du\), so \[ \frac{\sqrt{1+\ln x}}{x\ln x}\,dx =\frac{2u^2}{u^2-1}\,du =\left(2+\frac1{u-1}-\frac1{u+1}\right)du. \] Hence \[ \boxed{2\sqrt{1+\ln x}+\ln\left|\frac{\sqrt{1+\ln x}-1}{\sqrt{1+\ln x}+1}\right|+C.} \]

For the second, \[ \frac{7-\ln x}{3+\ln x}=-1+\frac{10}{3+\ln x}, \] so \[ \boxed{-\ln x+10\ln|3+\ln x|+C.} \]

For the third, \(u=1+\tan x\) gives \[ \boxed{\ln|1+\tan x|+C.} \]

For the fourth, \(u=\cos x/2\) gives \(du=-\tfrac12\sin x\,dx\) and \[ \boxed{-\arcsin\!\left(\frac{\cos x}{2}\right)+C.} \]

For the fifth, \(u=\tan x/3\) gives \(du=\tfrac13\sec^2x\,dx\) and \[ \boxed{\arcsin\!\left(\frac{\tan x}{3}\right)+C.} \]

Using \(1+\cos(2u)=2\cos^2u\), \[ \sqrt{1+\cos(2ax)}=\sqrt2\,|\cos(ax)|. \] This equals \(\sqrt2\cos(ax)\) only on intervals where \(\cos(ax)\ge0\).

If \(a\ne0\), on any interval on which \(\cos(ax)\) has constant sign, \[ \int \sqrt{1+\cos(2ax)}\,dx =\begin{cases} \dfrac{\sqrt2}{a}\sin(ax)+C,&\cos(ax)\ge0,\\[6pt] -\dfrac{\sqrt2}{a}\sin(ax)+C,&\cos(ax)\le0. \end{cases} \] These local primitives glue continuously across the zeros of \(\cos(ax)\) after adjusting the constants. Equivalently, a global primitive is \[ \frac{\sqrt2}{a}H(ax)+C, \] where, for \[ k=\left\lfloor\frac{u+\pi/2}{\pi}\right\rfloor, \qquad H(u)=2k+\sin(u-k\pi). \] Then \(H'(u)=|\cos u|\) wherever differentiated, including across the joining points by continuity of the derivative. If \(a=0\), the integrand is the constant \(\sqrt2\) and the primitive is \(\sqrt2 x+C\).

Direct expansion/substitution gives, respectively, \[ \boxed{x-\frac12\cos2x+C,} \] \[ \boxed{\sin^3x-\cos^3x+C,} \] \[ \boxed{-\frac13\cos^3x+\frac12(x+\sin x\cos x)+C,} \] \[ \boxed{-\frac16\cos^6x+\frac18\cos^8x+C,} \] \[ \boxed{-\frac14\cos^4x+C,} \] and, on intervals where \(\sqrt{\cos x}\) is real, \[ \boxed{-\frac23\cos^{3/2}x+\frac27\cos^{7/2}x+C.} \] For example, the fourth follows by writing \(\sin^3x=\sin x(1-\cos^2x)\) and taking \(u=\cos x\); the other odd-power cases are analogous.

Using the power-reduction identities and elementary substitutions, \[ \boxed{\int\sin^2x\,dx=\frac12(x-\sin x\cos x)+C,} \] \[ \boxed{\int\cos^2x\,dx=\frac12(x+\sin x\cos x)+C,} \] \[ \boxed{\int\sin^3x\,dx=\frac13\cos^3x-\cos x+C,} \] \[ \boxed{\int\cos^3x\,dx=\sin x-\frac13\sin^3x+C.} \] With \(u=\sqrt x\), the nested-radical variant is \[ \boxed{\int \frac{\sin^3\sqrt x}{2\sqrt x}\,dx =\frac13\cos^3\sqrt x-\cos\sqrt x+C.} \] Expanding the cubes gives \[ \boxed{\int(1+\sin x)^3dx =\frac52x+\frac13\cos^3x-4\cos x-\frac34\sin2x+C,} \] \[ \boxed{\int(1+\cos x)^3dx =\frac52x-\frac13\sin^3x+4\sin x+\frac34\sin2x+C.} \] Reduction formulas yield \[ \boxed{\int\sin^4x\,dx =\frac38x-\frac38\sin x\cos x-\frac14\sin^3x\cos x+C,} \] \[ \boxed{\int\cos^4x\,dx =\frac38x+\frac38\sin x\cos x+\frac14\sin x\cos^3x+C.} \] Finally \(\cos^4x-\sin^4x=\cos2x\), so \[ \boxed{\int(\cos^4x-\sin^4x)dx=\frac12\sin2x+C.} \]

<1>1. Evaluation of the base integral \(\int \ln(ax + b) \, dx\): <2>1. Substitute \(u = ax + b\), so \(du = a \, dx \implies dx = \frac{1}{a} \, du\). The integral becomes: \[ \int \ln(ax + b) \, dx = \frac{1}{a} \int \ln(u) \, du. \]

<2>2. Integrating by parts with \(U = \ln u, \, dV = du \implies dU = \frac{1}{u} du, \, V = u\): \[ \int \ln u \, du = u \ln u - \int 1 \, du = u \ln u - u + C. \]

<2>3. Substituting back \(u = ax + b\): \[ \int \ln(ax + b) \, dx = \frac{ax + b}{a} \ln(ax + b) - \frac{ax + b}{a} + C = \frac{ax + b}{a} \ln(ax + b) - x + C. \]

<1>2. Evaluation of sub-problems (1) through (4): <2>1. Sub-problem 1: \(\int \log_3(x) \, dx\) Using the change of base formula \(\log_3(x) = \frac{\ln x}{\ln 3}\): \[ \int \log_3(x) \, dx = \frac{1}{\ln 3} \int \ln x \, dx = \frac{x \ln x - x}{\ln 3} + C. \]

<2>2. Sub-problem 2: \(\int_1^e \ln(\sqrt{x}) \, dx\) Using the logarithm power rule \(\ln(\sqrt{x}) = \frac{1}{2} \ln x\): \[ \int_1^e \ln(\sqrt{x}) \, dx = \frac{1}{2} \int_1^e \ln x \, dx = \frac{1}{2} \Big[ x \ln x - x \Big]_1^e = \frac{1}{2} \Big( (e - e) - (0 - 1) \Big) = \frac{1}{2}(1) = \frac{1}{2}. \]

<2>3. Sub-problem 3: \(\int \ln(x^2 + 6x + 5) \, dx\) Factoring \(x^2 + 6x + 5 = (x + 1)(x + 5)\): \[ \int \ln(x^2 + 6x + 5) \, dx = \int \ln(x + 1) \, dx + \int \ln(x + 5) \, dx. \] Applying <1>1 to each term gives: \[ \big((x + 1)\ln(x + 1) - x\big) + \big((x + 5)\ln(x + 5) - x\big) + C = (x + 1)\ln(x + 1) + (x + 5)\ln(x + 5) - 2x + C. \]

<2>4. Sub-problem 4: \(\int \ln(x^2 - 1) \, dx\) Factoring \(x^2 - 1 = (x - 1)(x + 1)\): \[ \int \ln(x^2 - 1) \, dx = \int \ln(x - 1) \, dx + \int \ln(x + 1) \, dx = (x - 1)\ln(x - 1) + (x + 1)\ln(x + 1) - 2x + C. \]

<1>3. Q.E.D.

Take \(n \neq -1\) and \(x > 0\).

<1>1. Integrate by parts with \(u = \ln x\) and \(dv = x^n\,dx\). <2>1. \(du = \frac{1}{x}\,dx\) and \(v = \frac{x^{n+1}}{n+1}\).

<2>2. \(\int x^n \ln x\,dx = \frac{x^{n+1}}{n+1}\ln x - \int \frac{x^{n+1}}{n+1}\cdot \frac{1}{x}\,dx\).

<2>3. \(= \frac{x^{n+1}}{n+1}\ln x - \frac{1}{n+1}\int x^n\,dx\).

<2>4. \(= \frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} + C\).

<1>2. \(\int x^n \ln x\,dx = \frac{x^{n+1}}{n+1}\left(\ln x - \frac{1}{n+1}\right) + C\).

<1>3. Q.E.D.

For \(a\ne0\), \[ \frac{d}{dx}\left(\frac{\sqrt{x^2-a^2}}{a^2x}\right) =\frac1{x^2\sqrt{x^2-a^2}}, \] so \[ \boxed{\int\frac{dx}{x^2\sqrt{x^2-a^2}} =\frac{\sqrt{x^2-a^2}}{a^2x}+C.} \] Hence \[ \boxed{\int\frac{dx}{x^2\sqrt{x^2-4}}=\frac{\sqrt{x^2-4}}{4x}+C,} \] \[ \boxed{\int\frac{dx}{x^2\sqrt{x^2-9}}=\frac{\sqrt{x^2-9}}{9x}+C.} \] Likewise, \[ \boxed{\int\frac{dx}{x^2\sqrt{4x^2-1}} =\frac{\sqrt{4x^2-1}}{x}+C,} \] as direct differentiation verifies.

For the final integral, put \(u=\sqrt{x^2-16}\). Since \(u\,du=x\,dx\) and \(x^2=u^2+16\), \[ \int\frac{dx}{2x\sqrt{x^2-16}} =\frac12\int\frac{du}{u^2+16} =\boxed{\frac18\arctan\!\left(\frac{\sqrt{x^2-16}}4\right)+C.} \] This form avoids the branch conventions inherent in \(\sec^{-1}(x/4)\).

Let \(u=x^3\), so \(du=3x^2\,dx\). Therefore \[ \boxed{\int\frac{x^2}{1+x^6}\,dx=\frac13\arctan(x^3)+C.} \]

Put \(u=e^x\), so \(dx=du/u\). Then \[ \frac{1+u}{u(1-u)}=\frac1u+\frac2{1-u}. \] Hence \[ \int \frac{1+e^x}{1-e^x}\,dx =\ln u-2\ln|1-u|+C =\boxed{x-2\ln|1-e^x|+C}. \]

With \(u=x/3\), \(dx=3\,du\), \[ \int\frac{dx}{9+x^2} =\frac13\int\frac{du}{1+u^2} =\boxed{\frac13\arctan\!\left(\frac x3\right)+C.} \]

Let \(u=x^2+2x+3\), so \(du=2(x+1)\,dx\). Therefore \[ \boxed{\int\frac{x+1}{x^2+2x+3}\,dx=\frac12\ln(x^2+2x+3)+C.} \]

Where the integrand is defined, \[ \cot x\tan x=1, \] so \[ \int \sin x\cos x\,dx =\boxed{\frac12\sin^2x+C.} \] This is an antiderivative on each open interval of the domain \(\{x:\sin x\cos x\neq0\}\) of the integrand.

Let \(u=x^2/2\), so \(du=x\,dx\) and \[ \sqrt{4-x^4}=2\sqrt{1-u^2}. \] Therefore \[ \int\frac{x}{\sqrt{4-x^4}}\,dx =\frac12\int\frac{du}{\sqrt{1-u^2}} =\boxed{\frac12\arcsin\!\left(\frac{x^2}{2}\right)+C.} \]

Let \(u=x+1\), so \(x=u-1\) and \(du=dx\). Then \[ (u-1)u^{1/2}=u^{3/2}-u^{1/2}, \] so \[ \int x\sqrt{x+1}\,dx =\frac25u^{5/2}-\frac23u^{3/2}+C. \] Therefore \[ \boxed{\int x\sqrt{x+1}\,dx=\frac25(x+1)^{5/2}-\frac23(x+1)^{3/2}+C.} \]

Since \[ \frac{\sin x}{1+\cos x}=\tan\frac x2, \qquad \frac1{1+\cos x}=\frac12\sec^2\frac x2, \] the first integrand is the derivative of \(x\tan(x/2)\). Hence \[ \boxed{\int\frac{x+\sin x}{1+\cos x}\,dx=x\tan\frac x2+C.} \]

For the second, integrate by parts using \[ \frac{e^x}{(e^x+1)^2} =-\frac{d}{dx}\frac1{e^x+1}. \] Then \[ \int\frac{x e^x}{(e^x+1)^2}\,dx =-\frac{x}{e^x+1}+\int\frac{dx}{e^x+1}. \] Since \[ \int\frac{dx}{e^x+1}=x-\ln(e^x+1)+C, \] we obtain \[ \boxed{\int\frac{x e^x}{(e^x+1)^2}\,dx =\frac{x e^x}{e^x+1}-\ln(e^x+1)+C.} \]

<1>1. Integrate \(x/(x^4-16)\).

<1>2. Integrate \(x\arctan x/(1+x^2)^2\).

Split the integral: \[ \int\frac{4x^3}{x^4+1}\,dx+\int\frac{2x}{x^4+1}\,dx. \] The first is \(\ln(x^4+1)\). In the second, put \(u=x^2\), so \(du=2x\,dx\) and the denominator is \(u^2+1\). Thus \[ \boxed{\int\frac{4x^3+2x}{x^4+1}\,dx=\ln(x^4+1)+\arctan(x^2)+C.} \]

For the first integral, the standard hyperbolic/trigonometric substitution gives \[ \boxed{\int\frac{dx}{\sqrt{x^2+25}} =\ln\!\left(x+\sqrt{x^2+25}\right)+C.} \] The logarithm’s argument is positive for every real \(x\).

For the second, \[ \frac{d}{dx}\left(\frac{x}{\sqrt{1+x^2}}\right) =\frac1{(1+x^2)^{3/2}}, \] so \[ \boxed{\int\frac{dx}{(1+x^2)^{3/2}} =\frac{x}{\sqrt{1+x^2}}+C.} \]

For (1), let \(t=\sqrt x\). Then \(dx=2t\,dt\) and \[ 2t\frac{1-t}{1+t}=-2t+4-\frac4{t+1}. \] Hence \[ \boxed{-x+4\sqrt x-4\ln(1+\sqrt x)+C.} \]

For (2), put \(t=x^{1/6}\). Then \[ \frac{dx}{\sqrt x+2\sqrt[3]x}=\frac{6t^3}{t+2}\,dt, \] so polynomial division gives \[ \boxed{2\sqrt x-6\sqrt[3]x+24\sqrt[6]x-48\ln|\sqrt[6]x+2|+C.} \]

For (3), with the same \(t=x^{1/6}\), \[ \frac{dx}{\sqrt x-\sqrt[3]x}=\frac{6t^3}{t-1}\,dt, \] thus \[ \boxed{2\sqrt x+3\sqrt[3]x+6\sqrt[6]x+6\ln|\sqrt[6]x-1|+C.} \]

For (4), let \(t=x^{1/4}\). Then \[ \frac{dx}{\sqrt x+\sqrt[4]x}=\frac{4t^2}{t+1}\,dt, \] so \[ \boxed{2\sqrt x-4\sqrt[4]x+4\ln|\sqrt[4]x+1|+C.} \]

For (5), let \(t=\sqrt[3]x\). Since \(dx=3t^2dt\), \[ 3t^2\frac{t+1}{t-1}=3t^2+6t+6+\frac6{t-1}, \] whence \[ \boxed{x+3x^{2/3}+6x^{1/3}+6\ln|x^{1/3}-1|+C.} \]

For the first integral in (6), let \(t=\sqrt x\): \[ \boxed{\int\frac{dx}{(x+1)\sqrt x}=2\arctan\sqrt x+C.} \] For the second, let \(t=\sqrt{x-1}\): \[ \boxed{\int\frac{dx}{2x\sqrt{x-1}}=\arctan\sqrt{x-1}+C.} \]

For (7), let \(u=x+1\): \[ \boxed{\frac35(x+1)^{5/3}-\frac32(x+1)^{2/3}+C.} \]

For (8), let \(u=\sqrt{4x-1}\). Then \[ \boxed{2\arctan\sqrt{4x-1}+C.} \]

For (9), let \(u=\sqrt{25x^4-1}\). Since \(u^2+1=25x^4\), the transformed integrand is \(du/(1+u^2)\), giving \[ \boxed{\arctan\sqrt{25x^4-1}+C.} \]

For (10), let \(u=2\sqrt x\). Then \(du=dx/\sqrt x\), so \[ \boxed{\arcsin(2\sqrt x)+C.} \] On \(0<x<1/4\) this is equivalent, up to an additive constant, to \(\tfrac12\arcsin(8x-1)\).

For (11), split the integral: \[ \int\frac6{\sqrt{9-x^2}}\,dx-\int\frac{2x}{\sqrt{9-x^2}}\,dx, \] which gives \[ \boxed{6\arcsin\!\left(\frac x3\right)+2\sqrt{9-x^2}+C.} \]

For (12), let \(u=x^{1/4}\). Since \[ \sqrt{x\sqrt x-x^2}=x^{3/4}\sqrt{1-\sqrt x}, \] and \(dx=4u^3du\), we obtain \[ \boxed{4\arcsin(x^{1/4})+C.} \]

Since \(e^{x+e^x}=e^x e^{e^x}\), let \(u=e^x\). Then \(du=e^x\,dx\), so \[ \boxed{\int e^{x+e^x}\,dx=e^{e^x}+C.} \]

For \(x>1\), \[ \frac{d}{dx}\sec^{-1}x=\frac1{x\sqrt{x^2-1}}, \qquad \frac{d}{dx}\csc^{-1}x=-\frac1{x\sqrt{x^2-1}}. \] Integration by parts gives \[ \boxed{\int x\sec^{-1}x\,dx =\frac12\left(x^2\sec^{-1}x-\sqrt{x^2-1}\right)+C,} \] and \[ \boxed{\int x\csc^{-1}x\,dx =\frac12\left(x^2\csc^{-1}x+\sqrt{x^2-1}\right)+C.} \]

1. We first need the derivatives of \(\arctan\), since \[ T(p, x) = \sum_{n=0}^\infty \frac{f^{(n)}(p)}{n!}(x-p)^n \implies T(0, x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n \]

   Starting with \(y=\arctan x\), we can write \(\tan y - x = 0 \implies \sec^2y~dy - 1~dx = 0\) by implicit differentiation and the chain rule, and so \(\dd{y}{x} = y' = \sec^{-2}y = \cos^2(\arctan x))\). Embed this in a triangle with angle \(y\) to find that \[y' = (1+x^2)^{-1}.\]

   Computing higher derivatives of \(\arctan\) directly is tedious, so instead we’ll take a power series of \(y'\) and differentiate that instead:

   \[\begin{align*} f'(x) = \inverseof{(1+x^2)} & =1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots = \sum_{i=0} (-1)^{i}x^{2i} \\ \implies f''(x) = \dd{}{x} \inverseof{(1+x^2)} &= -2x + 4x^3 - 6x^5 + \cdots = \sum_{i=1}^\infty (2i)(-1)^{i}x^{2i-1} \\ \implies f^{(3)}(x) = \dd{^2}{x^2} \inverseof{(1+x^2)} &= -2 + 12x^2 - 30x^4 + \cdots = \sum_{i=1}^\infty (2i )(2i-1)(-1)^{i}x^{2i-2} \\ \implies f^{(4)}(x) = \dd{^3}{x^3} \inverseof{(1+x^2)} &= 24x - 120x^3 + \cdots = \sum_{i=2}^\infty (2i)(2i-1)(2i-2)(-1)^{i}x^{2i-3} \\ \implies f^{(5)}(x) = \dd{^4}{x^4} \inverseof{(1+x^2)} &= 24 - 360x^2 + \cdots = \sum_{i=2}^\infty (2i)(2i-1)(2i-2)(2i-3)(-1)^{i}x^{2i-4} \\ \end{align*}\]

   so recalling that we’ll be evaluating at \(x=0\), we can notice that whenever \(n+1\) is even (and thus \(n\) is odd), there is no constant term. So for \(n+1\) odd, we can extract the general pattern: \[\begin{align*} f^{(n)}(0) &= \restrictionof{ \dd{^{n}}{x^{n}} \inverseof{(1+x^2)}}{x = 0} &= \begin{cases} (n-1)! & n-1 \equiv 0 \mod 4 \\ -(n-1)! & n -1 \equiv 2 \mod 4 \end{cases} \end{align*}\]

   Thus the desired Taylor expansion will be \[ T(0, x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n \\ = \arctan(0) + \sum_{n>0\text{ odd}} \sigma_n \frac{(n-1)!}{n!} x^n \\ =\sum_{n>0\text{ odd}} \frac{\sigma_nx^n}{n} \]

   where \(\sigma_n = -1 \iff n \equiv 3 \mod 4\).

   We can then start writing this out: \[ T(0, x) = x - \frac 1 3 x^3 + + \frac{1}{5}x^5 - \frac 1 7 x^7 \]

   We then want an \(n\) such that \[\frac{(\frac{n-1}{2})!}{(n-1)!} (\frac 1 {10^2})^n \leq \frac 1 {10^5}.\]

   Checking manually, \(i=1\) does not work, but \(i=2\) does, so we take the approximation \[\widehat f(x) = x - \frac 1 3 x^3.\] \(\qed\)

Goal: Derive the Maclaurin series expansion of \(f(x) = \arctan(x)\) and determine a polynomial approximation accurate within \(10^{-5}\) for small \(|x| \le 10^{-2}\).

<1>1. For all \(x \in (-1, 1)\), \(f'(x) = \frac{1}{1+x^2} = \sum_{k=0}^\infty (-1)^k x^{2k}\).

<1>2. The Maclaurin series of \(\arctan(x)\) is \(T(0, x) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} x^{2k+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\) for \(|x| < 1\).

<1>3. The derivatives of \(f\) at \(x=0\) satisfy: \[f^{(n)}(0) = \begin{cases} 0 & \text{if } n \text{ is even}, \\ (-1)^k (2k)! & \text{if } n = 2k+1 \text{ is odd}.\end{cases}\]

<1>4. For \(|x| \le 10^{-2}\), the polynomial approximation \(\widehat{f}(x) = x - \frac{x^3}{3}\) approximates \(\arctan(x)\) with error bounded by \(\frac{1}{5 \cdot 10^{10}} < 10^{-5}\).

All identities follow from \[ \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B, \qquad \sin(A\pm B)=\sin A\cos B\pm\cos A\sin B. \]

Adding the two cosine formulas gives \[ \cos(A+B)+\cos(A-B)=2\cos A\cos B. \] With \(A=(x+y)/2\) and \(B=(x-y)/2\) this becomes \[ \cos x+\cos y =2\cos\frac{x+y}{2}\cos\frac{x-y}{2}. \] Subtracting the same formulas gives \[ \cos(A+B)-\cos(A-B)=-2\sin A\sin B, \] so \[ \cos x-\cos y =-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}. \]

When \(\cos x\cos y\ne0\), \[ \tan x-\tan y =\frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y} =\frac{\sin(x-y)}{\cos x\cos y}. \]

Adding and subtracting the sine formulas yields \[ \sin(A+B)+\sin(A-B)=2\sin A\cos B, \] \[ \sin(A+B)-\sin(A-B)=2\cos A\sin B. \] Therefore \[ \sin x\cos y=\frac12\bigl(\sin(x+y)+\sin(x-y)\bigr), \] \[ \cos x\sin y=\frac12\bigl(\sin(x+y)-\sin(x-y)\bigr). \] Finally, adding and subtracting the cosine formulas gives \[ \cos x\cos y=\frac12\bigl(\cos(x+y)+\cos(x-y)\bigr), \] \[ \sin x\sin y=\frac12\bigl(\cos(x-y)-\cos(x+y)\bigr). \]