Question
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$

Answer

Let $\sin^2\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$2\sin\text{x}\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
$=\frac{1}{2}\int\limits^1_0\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\big[\tan^{-1}\text{t}\big]^1_0$
$=\frac{1}{2}\Big[\tan^{-1}(1)-\tan^{-1}(0)\Big]$
$=\frac{1}{2}\Big[\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)-\tan^{-1}(\tan0)\Big]$
$=\frac{1}{2}\times\frac{\pi}{4}$
$=\frac{\pi}{8}$ 

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