TS EAMCET · Maths · Definite Integration
\(\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=\)
- A \(\frac{\pi}{4}+\frac{2}{3} \tan ^{-1} 2\)
- B \(-\frac{\pi}{3}-\frac{2}{3} \tan ^{-1} 3\)
- C \(-\frac{\pi}{12}+\frac{2}{3} \tan ^{-1} 2\)
- D \(\frac{\pi}{6}-\frac{2}{3} \tan ^{-1} 4\)
Answer & Solution
Correct Answer
(C) \(-\frac{\pi}{12}+\frac{2}{3} \tan ^{-1} 2\)
Step-by-step Solution
Detailed explanation
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…
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