WBJEE · Maths · Definite Integration
The value of the integral \(\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x\)
is equal to
- A \(\log _{e} 2\)
- B \(\log _{e} 3\)
- C \(\frac{1}{4} \log _{e} 2\)
- D \(\frac{1}{4} \log _{e} 3\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4} \log _{e} 3\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x\) \(=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+2 \sin x \cos x} d x\) \(=\int_{0}^{\pi / 4}-\frac{\sin x+\cos x}{(\sin x-\cos x)^{2}-4} d x\) Put \(\sin x-\cos x=t\) \(\Rightarrow \quad(\cos x+\sin x) d x=d t\) when…
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