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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Definite Integration

The integral ∫_0^π ((x+3) x)/(1+3 ^2 x) d x is equal to :

The integral ∫0π(x+3)sin⁡x1+3cos⁡2xdx\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x is equal to :

Options

  1. Aπ3(π+1)\frac{\pi}{\sqrt{3}}(\pi+1)
  2. Bπ3(π+2)\frac{\pi}{\sqrt{3}}(\pi+2)
  3. Cπ33(π+6)\frac{\pi}{3 \sqrt{3}}(\pi+6)
  4. Dπ23(π+4)\frac{\pi}{2 \sqrt{3}}(\pi+4)

Correct answer

C. π33(π+6)\frac{\pi}{3 \sqrt{3}}(\pi+6)

Step-by-step solution

I=∫0π(x+3)sin⁡x1+3cos⁡2xdxI=∫0π(π−x+3)sin⁡x(1+3cos⁡2x)dx2I=∫0π(π+6)sin⁡x⋅dx(1+3cos⁡2x)=2∫0π/2(π+6)sin⁡x(1+3cos⁡2x)I=∫0π/2(π+6)sin⁡x⋅dx(1+3cos⁡2x)=π33(π+6)3cos⁡x=t3sin⁡x=dt\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 ^2 x\right)} \\ & I=\int_0^{\pi / 2} \frac{(\pi+6) \sin x \cdot d x}{\left(1+3 \cos ^2 x\right)}=\frac{\pi}{3 \sqrt{3}}(\pi+6) \\ & \sqrt{3} \cos x=t \\ & \sqrt{3} \sin x=d t\end{aligned}
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