Evaluate \[ \int\sqrt{1-x^2}\,dx, \qquad \int_{-4}^{4}\sqrt{16-x^2}\,dx, \] and \[ \int_{-1}^{1}(1+x)\sqrt{1-x^2}\,dx. \]
Solution
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 \[ \boxed{\int_{-1}^{1}(1+x)\sqrt{1-x^2}\,dx=\frac\pi2.} \]