Mathematics

$$\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$$


SOLUTION

Let $$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$$.

Then

$$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}dx$$. 

Then, 

$$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sec^{2}x}{\sec^{2}x(\sec^{2}x+3\tan^{2}x)}\ dx$$                       [Diving $$N^{r}$$ and $$D^{r}$$ by $$\cos^{4}x$$]

$$\Rightarrow I=\displaystyle\int_{0}^{\infty}\dfrac{1}{(1+t^{2})(1+4t^{2})}\ dt$$, 

where $$t=\tan x$$

$$\Rightarrow I=-\dfrac{1}{3}\displaystyle\int_{0}^{\infty}\left(\dfrac{1}{1+t^{2}}-\dfrac{4}{1+t^{2}}\right)dt$$ 

$$-\dfrac{1}{3}\left[\tan^{-1}t-2\tan^{-1}2t\right]_{0}^{\infty}$$

$$\Rightarrow =-\dfrac{1}{3}\left(\dfrac{\pi}{2}-\pi\right)=\dfrac{\pi}{6}$$

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Subjective Medium Published on 17th 09, 2020
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