2015 Cambridge Entrance Examination Q6 
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*Show : $\sec^2\left(\frac{\pi}{4}-\frac{1}{2}x\right)=\frac{2}{1+\sin(x)}\\$. Then evaluate$\int\frac{dx}{1+\sin(x)}$

*Show : $\int_{0}^{\pi}x \space f(\sin(x)) \space dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x)) \space dx$. Then evaluate $\int_{0}^{\pi}\frac{x}{1+\sin(x)}$

*Evaluate: $\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin(x))^2}dx$
The first trigonometric identity is quite easy to prove.
$$\sec^2\left(\frac{\pi}{4}-\frac{1}{2}x\right)=\frac{1}{\cos^2\left(\frac{1}{2}\left((\frac{\pi}{2}-x\right)\right)}=\frac{1}{1/2\left(1+\cos\left(\frac{\pi}{2}-x\right)\right)}=\frac{2}{1+\sin(x)}$$
And the integral just involves substitution the integrand with the secant term, which alloys us to take advantage of the fact that the derivative of the tangent is secant squared.
The first second integral were tricky at first but allowing $y=\pi-x$ did the trick. However evaluting the integral in the third part...
$$\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin(x))^2}$$
... wasn't as simple as I thought. I'm really not sure where to begin or how to utelise the integral properties proved in the previous parts. Thanks
 A: The beauty of this problem is that it used an allied function $\sin(x)$ with phase $\pi$
If you could see the graph of $y = E(x)\times A(x)$ it can help you understand the complexity of the problems
Basically if $E(x)$ is an odd function then this can be solved:

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*Solution
$\int_0^\pi x^nf(\sin x)dx$ can be solved if $n $ is an odd
Given: $\int_0^\pi x f(\sin x) dx = \frac \pi2\int_0^\pi f(\sin x)dx$

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*To find $\int_0^\pi \frac {2x^3-3\pi x^2}{(1+\sin x)^2}dx = \int_0^\pi (2x^3-3\pi x^2) f(\sin x)$
Take
$$\begin{align*}
\int_0^\pi x^3 f(\sin x)dx
& = \int_0^\pi(\pi-x)^3f(\sin x)dx\\
& \text { by expansion.......and using the given property }\\
& \implies \int_0^\pi x^3f(\sin x)dx = \int_0^\pi \left(\frac {\pi^3}{2}-\frac{3\pi^3}4 + \frac {3\pi x^2}{2} \right)f(\sin x)dx
\end{align*}$$
Coming back to our original problem:
$$I = \int_0^\pi \frac {2\left(-\frac {\pi^3}{2} + \color{blue}{\frac {3\pi x^2}{2}}\right) - \color{blue}{3\pi x^2}}{(1+\sin x)^2}dx$$
Now, I believe you can solve $\int_0^\pi\text {constant }\frac {1}{(1+\sin x)^2}dx$
