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Question: Answered & Verified by Expert
$\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=$
MathematicsDefinite IntegrationTS EAMCETTS EAMCET 2019 (06 May Shift 1)
Options:
  • A $\frac{\pi}{4}+\frac{2}{3} \tan ^{-1} 2$
  • B $-\frac{\pi}{3}-\frac{2}{3} \tan ^{-1} 3$
  • C $-\frac{\pi}{12}+\frac{2}{3} \tan ^{-1} 2$
  • D $\frac{\pi}{6}-\frac{2}{3} \tan ^{-1} 4$
Solution:
1675 Upvotes Verified Answer
The correct answer is: $-\frac{\pi}{12}+\frac{2}{3} \tan ^{-1} 2$
We have,
$\int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x$
$=\int_0^{\pi / 4} \frac{d x}{1+4 \tan ^2 x}$
put $\tan x=t$
$\begin{aligned} & \Rightarrow \sec ^2 x d x=d t \\ & \Rightarrow d x=\frac{d t}{\sec ^2 x}=\frac{d t}{1+t^2}\end{aligned}$
when $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$
From Eq. (i), $\int_0^1 \frac{d t}{\left(1+t^2\right)\left(1+4 t^2\right)}$
$\begin{aligned} & =\frac{1}{3} \int_0^1 \frac{4\left(1+t^2\right)-\left(1+4 t^2\right)}{\left(1+4 t^2\right)\left(1+t^2\right)} d x \\ & =\frac{1}{3} \int_0^1\left(\frac{4}{1+4 t^2}-\frac{1}{1+t^2}\right) d t \\ & =\frac{1}{3}\left[\frac{4}{4} \int_0^1 \frac{d t}{\left(\frac{1}{2}\right)^2+t^2}-\int_0^1 \frac{1}{1+t^2} d t\right] \\ & =\frac{1}{3}\left[\frac{1}{1 / 2}\left(\tan ^{-1} \frac{t}{1 / 2}\right)_0^1-\left(\tan ^{-1} t\right)_0^1\right] \\ & =\frac{1}{3}\left[2\left(\tan ^{-1} 2 t\right)_0^1-\frac{\pi}{4}\right] \\ & =\frac{1}{3}\left[2 \tan ^{-1}\left(2-\frac{\pi}{4}\right]=-\frac{\pi}{12}+\frac{2}{3} \tan ^{-1} 2\right.\end{aligned}$

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