Integration #6

If we substitute $y = \sin x$ and then integrate by parts twice, we obtain $\begin{aligned} \int_0^{\frac12\pi} \frac{x^2 \cos x}{(1 + \sin x)^2}\,dx & = \; \int_0^1 \frac{(\sin^{-1}y)^2}{(1+y)^2}\,dy \; = \; \left[-\frac{(\sin^{-1}y)^2}{1+y}\right]_0^1 + 2\int_0^1 \frac{\sin^{-1}y}{\sqrt{1-y^2}(1+y)}\,dy \\ & = \; -\tfrac18\pi^2 + 2\int_0^1 \frac{\sin^{-1}y}{(1-y)^{\frac12}(1+y)^{\frac32}}\,dy \; = \; -\tfrac18\pi^2 + 2\left[-\sin^{-1}y\,\sqrt{\frac{1-y}{1+y}}\right]_0^1 + 2\int_0^1 \frac{1}{\sqrt{1-y^2}}\,\sqrt{\frac{1-y}{1+y}}\,dy \\ & = \; 2\int_0^1 \frac{dy}{1+y} - \tfrac18\pi^2 \; = \; 2\ln2 - \tfrac18\pi^2 \end{aligned}$ making the answer $\boxed{0.152593811}$ .

For integration by parts, let $u = x^2, du= 2x dx, dv = \frac{ \cos x dx}{(1 + \sin x)^2}$ . Then $v = \int \frac{ \cos x dx}{(1 + \sin x)^2}$ . To quickly compute this antiderivative let $w = 1 + \sin x$ so $dw = \cos x dx$ so the integral becomes $\int \frac{dw}{w^2} = -w^{-1} = -\frac{1}{1 + \sin x}$ . Then, the whole integral is equal to:

$-\frac{x^2}{1 + \sin x} \bigg|_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \frac{2xdx}{1 + \sin x} = -\frac{\pi^2}{8} + \int_0^{\frac{\pi}{2}} \frac{2xdx}{1 + \sin x}$ . The remaining integral is very tricky but it can be turned into a known integral by playing with it a little:

$1 + \sin x = 1 + \cos (\frac{\pi}{2} - x) = 1 + \cos( 2( \frac{\pi}{4} - \frac{x}{2} )) = 2 \cos^2 ( \frac{\pi}{4} - \frac{x}{2})$ so the integral is $\int_0^{\frac{\pi}{2}} x \sec^2 ( \frac{\pi}{4} - \frac{x}{2} ) dx$ . Finally, with the substitution $u = \frac{\pi}{4} - \frac{x}{2}$ the integral becomes $\int_0^{\frac{\pi}{4}} (\pi - 4u) \sec^2 u du$ . This reduces the problem to computing $\int \sec^2 x dx$ and $\int x \sec^2 x dx$ which are standard integrals. In the end, it evaluates to $\log(4)$ so the solution is $\log(4) - \frac{\pi^2}{8}$ .