TS EAMCET · Maths · Definite Integration
\(\int_0^{\frac{\pi}{4}} \frac{\sec x}{1+2 \sin ^2 x} d x=\)
- A \(\frac{1}{3} \log (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{12}\)
- B \(\frac{2}{3} \log (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{6}\)
- C \(\frac{1}{6} \log (\sqrt{2}-1)+\frac{\pi}{12}\)
- D \(\frac{1}{4} \log (\sqrt{2}-1)-\frac{\pi \sqrt{3}}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3} \log (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{12}\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \int_0^{\pi / 4} \frac{\sec x}{1+2 \sin ^2 x} d x \\ = & \int_0^{\pi / 4} \frac{\cos x}{\cos ^2 x\left(1+2 \sin ^2 x\right)} d x \\ = & \int_0^{\pi / 4} \frac{\cos x d x}{\left(1-\sin ^2 x\right)\left(1+2 \sin ^2 x\right)} d x \end{aligned} \] Let…
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