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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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