TS EAMCET · Maths · Definite Integration
\(\int_0^{\pi / 4} \frac{1}{5 \cos ^2 x+16 \sin ^2 x+8 \sin x \cos x} d x=\)
- A \(\operatorname{Tan}^{-1}\left(\frac{4}{5}\right)\)
- B \(2 \operatorname{Tan}^{-1}\left(\frac{3}{5}\right)\)
- C \(\frac{1}{8} \operatorname{Tan}^{-1}\left(\frac{8}{9}\right)\)
- D \(\frac{1}{4} \operatorname{Tan}^{-1}\left(\frac{7}{8}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{8} \operatorname{Tan}^{-1}\left(\frac{8}{9}\right)\)
Step-by-step Solution
Detailed explanation
\(\int_0^{\pi / 4} \frac{\sec ^2 x}{5 + 16 \tan ^2 x + 8 \tan x} d x\) Let \(u = \tan x\), \(du = \sec ^2 x d x\). Limits: \(u(0)=0\), \(u(\pi/4)=1\). \(\int_0^1 \frac{1}{16 u^2 + 8 u + 5} d u\)…
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