# Thread: Having trouble evaluating an integral

1. ## Having trouble evaluating an integral

How do I go from the step indicated to the other step (Red arrow)? The red arrow points to the first part that's incorrect according to wolfram alpha.

My work is attached.

Any help in catching my mistake would be GREATLY appreciated!

P.S.
The answer I am supposed to get is 4*pi.

2. ## Re: Having trouble evaluating an integral

Originally Posted by s3a
How do I go from the step indicated to the other step (Red arrow)? The red arrow points to the first part that's incorrect according to wolfram alpha.

My work is attached.

Any help in catching my mistake would be GREATLY appreciated!

P.S.
The answer I am supposed to get is 4*pi.
\displaystyle \begin{align*} 4\int_0^{\sqrt{2}}{4\sqrt{2 - x^2} - 2x^2\sqrt{2 - x^2} - \frac{2}{3}\left(2 - x^2\right)^{\frac{3}{2}}\,dx} &= 4\int_0^{\sqrt{2}}{2\sqrt{2 - x^2}\left(2 - x^2\right) - \frac{2}{3}\left(2 - x^2\right)^{\frac{3}{2}}\,dx} \\ &= 4\int_0^{\sqrt{2}}{2\left(2 - x^2\right)^{\frac{3}{2}} - \frac{2}{3}\left(2 - x^2\right)^{\frac{3}{2}}\,dx} \\ &= 4\int{\frac{4}{3}\left(2 - x^2\right)^{\frac{3}{2}}\,dx} \\ &= \frac{16}{3}\int_0^{\sqrt{2}}{\left(2 - x^2\right)^{\frac{3}{2}}\,dx}\end{align*}

Now let \displaystyle \begin{align*} x = \sqrt{2}\sin{\theta} \implies dx = \sqrt{2}\cos{\theta}\,d\theta \end{align*} and note that when \displaystyle \begin{align*} x = 0, \theta = 0 \end{align*} and when \displaystyle \begin{align*} x = \sqrt{2}, \theta = \frac{\pi}{2} \end{align*} and the integral becomes

\displaystyle \begin{align*} \frac{16}{3}\int_0^{\sqrt{2}}{\left(2 - x^2\right)^{\frac{3}{2}}\,dx} &= \frac{16}{3}\int_0^{\frac{\pi}{2}}{\left[2 - \left(\sqrt{2}\sin{\theta}\right)^2\right]^{\frac{3}{2}}\,\sqrt{2}\cos{\theta}\,d\theta} \\ &= \frac{16\sqrt{2}}{3}\int_0^{\frac{\pi}{2}}{\left(2 - 2\sin^2{\theta}\right)^{\frac{3}{2}}\,\cos{\theta} \,d\theta} \\ &= \frac{16\sqrt{2}}{3}\int_0^{\frac{\pi}{2}}{\left[2\left(1 - \sin^2{\theta}\right)\right]^{\frac{3}{2}}\,\cos{\theta}\,d\theta} \\ &= \frac{16\sqrt{2}}{3}\int_0^{\frac{\pi}{2}}{\left(2 \cos^2{\theta}\right)^{\frac{3}{2}}\,\cos{\theta} \, d\theta} \\ &= \frac{16\sqrt{2}}{3}\int_0^{\frac{\pi}{2}}{2\sqrt{ 2}\cos^3{\theta}\cos{\theta}\,d\theta} \\ &= \frac{64}{3}\int_0^{\frac{\pi}{2}}{\cos^4{\theta} \, d\theta} \\ &= \frac{64}{3}\int_0^{\frac{\pi}{2}}{\left(\frac{1}{ 2} + \frac{1}{2}\cos{2\theta}\right)^2\,d\theta} \end{align*}

\displaystyle \begin{align*} &= \frac{64}{3}\int_0^{\frac{\pi}{2}}{\frac{1}{4} + \frac{1}{2}\cos{2\theta} + \frac{1}{4}\cos^2{2\theta}\,d\theta} \\ &= \frac{64}{3} \int_0^{\frac{\pi}{2}}{ \frac{1}{4} + \frac{1}{2} \cos{ 2 \theta } + \frac{1}{4}\left( \frac{1}{2} + \frac{1}{2} \cos{ 4\theta } \right) \,d\theta} \\ &= \frac{64}{3} \int_0^{\frac{\pi}{2}}{\frac{1}{4} + \frac{1}{2}\cos{2\theta} + \frac{1}{8} + \frac{1}{8}\cos{4\theta}\,d\theta} \\ &= \frac{64}{3}\int_0^{\frac{\pi}{2}}{\frac{3}{8} + \frac{1}{2}\cos{2\theta} + \frac{1}{8}\cos{4\theta}\,d\theta} \\ &= \frac{8}{3}\int_0^{\frac{\pi}{2}}{3 + 4\cos{2\theta} + \cos{4\theta}\,d\theta} \\ &= \frac{8}{3}\left[3\theta + 2\sin{2\theta} + \frac{1}{4}\sin{4\theta}\right]_0^{\frac{\pi}{2}} \\ &= \frac{8}{3}\left[\left(\frac{3\pi}{2} + 2\sin{\pi} + \frac{1}{4}\sin{2\pi} \right) - \left(3 \cdot 0 + 2\sin{0} + \frac{1}{4}\sin{0} \right)\right] \\ &= 4\pi \end{align*}

Thank you!