- \(\displaystyle \int \frac {\sec^3(x) + e^{\sin(x)}}{\sec(x)} ~dx = \color {blue} {\tan(x) + e^{\sin(x)}}\)
Solution: \(\frac {\sec ^3 (x) + e^{\sin(x)}}{\sec(x)} = \sec ^2 (x) + \cos (x) \cdot e^{\sin(x)}\)
Used 2018
- \(\displaystyle \int \frac {1}{\tan (x) + \cot (x)} ~dx = \color {blue} {\frac {1}{2}\sin ^2 (x)} = \color {blue} {- \frac {1}{2}\cos ^2 (x)}\)
- Solution: \(\frac {1}{\tan (x) + \cot (x)} = \frac {1}{\frac {\sin (x)}{\cos (x)} + \frac {\cos (x)}{\sin (x)}} = \frac {\sin (x) \cos (x)}{\sin ^2 (x) + \cos ^2 (x)} = \sin (x) \cos (x)\)
Solution
1. \(\int \frac{\sec^3(x) + e^{\sin(x)}}{\sec(x)} \, dx = \tan(x) + e^{\sin(x)} + C\).
1.1. Simplify the integrand algebraically using \(\frac{1}{\sec(x)} = \cos(x)\): \[\frac{\sec^3(x) + e^{\sin(x)}}{\sec(x)} = \sec^2(x) + \cos(x) e^{\sin(x)}.\]
1.2. Integrate term-by-term: \[\int \sec^2(x) \, dx = \tan(x) + C_1.\]
1.3. For \(\int \cos(x) e^{\sin(x)} \, dx\), substitute \(u = \sin(x) \implies du = \cos(x) \, dx\): \[\int \cos(x) e^{\sin(x)} \, dx = \int e^u \, du = e^u + C_2 = e^{\sin(x)} + C_2.\]
1.4. Combining yields \(\tan(x) + e^{\sin(x)} + C\).
2. \(\int \frac{1}{\tan(x) + \cot(x)} \, dx = \frac{1}{2}\sin^2(x) + C = -\frac{1}{2}\cos^2(x) + C'\).
2.1. Rewrite trigonometric functions in terms of \(\sin(x)\) and \(\cos(x)\): \[\tan(x) + \cot(x) = \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} = \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)} = \frac{1}{\sin(x)\cos(x)}.\]
2.2. Thus the integrand is \(\frac{1}{\tan(x) + \cot(x)} = \sin(x)\cos(x)\).
2.3. Using the substitution \(u = \sin(x) \implies du = \cos(x) \, dx\): \[\int \sin(x)\cos(x) \, dx = \int u \, du = \frac{1}{2} u^2 + C = \frac{1}{2}\sin^2(x) + C.\]
2.4. Using \(\sin^2(x) = 1 - \cos^2(x)\), this equals \(-\frac{1}{2}\cos^2(x) + C'\).