AP EAMCET · Maths · Indefinite 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
\(\begin{aligned} & \text {Let } I=\int \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x \\ & =\int \frac{\cos ^2 x}{3 \sin ^2 x+1} d x=\int \frac{\sec ^2 x}{\left(\tan ^2 x+1\right)\left(4 \tan ^2 x+1\right)} d x \end{aligned}\) Let \(u=\tan x \Rightarrow d u=\sec ^2 x d x\)…
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