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

Answer

Let $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{1+\sin^2\text{x}}\text{ dx}$
Let $\sin\text{x}=\text{t}$ Then, $\cos\text{x dx}=\text{dt}$
When, $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{1+\sin^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{1}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=\frac{\pi}{4}$

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