Mathematics

Evaluate :
$$\displaystyle{{\int}_{0}}^{\pi /2} \dfrac{\sin x}{1+cos^2x}dx$$


SOLUTION
$$I={{\int}_{0}}^{\pi /2} \dfrac{\sin x }{1+cos^2x}dx$$

Let $$\cos x =t$$
$$-\sin x dx=dt$$
Now, $$x=0 \implies t=1$$ and $$x=\dfrac{\pi}{2} \implies t=0$$

Therefore,
$$I={{\int}_{1}}^0 \dfrac{\sin x}{1+t^2}(\dfrac{-dt}{\sin x})$$

$$I=-{{\int}_{1}}^0 \dfrac{dt}{1+t^2}$$

$$I=-{[tan^{-1}t]_{1}}^0$$

$$I=-[tan^{-1}0-tan^{-1}1]=-[0-\dfrac{\pi}{4}]=\dfrac{\pi}{4}$$
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Subjective Medium Published on 17th 09, 2020
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