JEE Mains · Maths · STD 12 - 7.2 definite integral
The integral \(80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16 \sin 2 \theta}\right) d \theta\) is equal to :
- A \(3 \log _e 4\)
- B \(4 \log _{\mathrm{e}} 3\)
- C \(6 \log _e 4\)
- D \(2 \log _e 3\)
Answer & Solution
Correct Answer
(B) \(4 \log _{\mathrm{e}} 3\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & I=\int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9-16 \sin 2 \theta}\right) d \theta \\ & \text { Take } \sin \theta-\cos \theta=t \\ & (\cos \theta+\sin \theta) d \theta=d t \\ & (\sin \theta-\cos \theta)^2=t^2 \\ & \Rightarrow \sin 2 \theta=1-t^2…
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