JEE Mains · Maths · STD 12 - 7.2 definite integral
The integral \(\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x\) is equal to :
- A \(\frac{\pi}{\sqrt{3}}(\pi+1)\)
- B \(\frac{\pi}{\sqrt{3}}(\pi+2)\)
- C \(\frac{\pi}{3 \sqrt{3}}(\pi+6)\)
- D \(\frac{\pi}{2 \sqrt{3}}(\pi+4)\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{3 \sqrt{3}}(\pi+6)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & I=\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x \\ & I=\int_0^\pi \frac{(\pi-x+3) \sin x}{\left(1+3 \cos ^2 x\right)} d x \\ & 2 I=\int_0^\pi \frac{(\pi+6) \sin x \cdot d x}{\left(1+3 \cos ^2 x\right)}=2 \int_0^{\pi / 2} \frac{(\pi+6) \sin x}{\left(1+3 \cos…
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