Q111 / 160 · AP EAMCET 2023 (15 May Shift 2) · Mathematics
$\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=$
- A$\frac{\pi}{4}+\frac{2}{3} \tan ^{-1} 2$
- B$-\frac{\pi}{3}-\frac{2}{3} \tan ^{-1} 3$
- Correct answerC$-\frac{\pi}{12}+\frac{2}{3} \tan ^{-1} 2$
- D$\frac{\pi}{6}-\frac{2}{3} \tan ^{-1} 4$
Solution
$\begin{aligned}
& \text {Let } I=\int \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x \\
& =\int \frac{\cos ^2 x}{3 \sin ^2 x+1} d x=\int \frac{\sec ^2 x}{\left(\tan ^2 x+1\right)\left(4 \tan ^2 x+1\right)} d x
\end{aligned}$
Let $u=\tan x \Rightarrow d u=\sec ^2 x d x$
$\begin{aligned}
& I=\int \frac{d u}{\left(u^2+1\right)\left(4 u^2+1\right)}=\int\left(\frac{4 / 3}{4 u^2+1}-\frac{1 / 3}{u^2+1}\right) d u \\
& \frac{2}{3} \tan ^{-1}(2 u)-\frac{1}{3} \tan ^{-1}(u)+c
\end{aligned}$
Now $\int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=\frac{2}{3} \tan ^{-1}(2)-\frac{\pi}{12}$
Hence, the correct answer is option: 'C'.