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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Definite Integration

NDA 2010 · Mathematics · Definite Integration

What is ∫_0^ π / 2 ( ^2 x)/( ^3 x+ ^3 x) d x ?

What is ∫0π/2sin⁡2xsin⁡3x+cos⁡3xdx\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin ^3 x+\cos ^3 x} d x ?

Options

  1. Aπ\pi
  2. Bπ2\frac{\pi}{2}
  3. Cπ4\frac{\pi}{4}
  4. D0

Correct answer

C. π4\frac{\pi}{4}

Step-by-step solution

I=∫0π/2sin⁡3xsin⁡3x+cos⁡2xdx\mathrm{I}=\int_0^{\pi / 2} \frac{\sin ^3 x}{\sin ^3 x+\cos ^2 x} d x...(i) I=∫0π/2{sin⁡(π2−x1)}3{sin⁡(π2−x)+}3+{cos⁡(π2−x)}3dxdxI=∫0π/2cos⁡3xsin⁡3x+cos⁡3xdx\begin{aligned} & \mathrm{I}=\int_0^{\pi / 2} \frac{\left\{\sin \left(\frac{\pi}{2}-x_1\right)\right\}^3}{\left\{\sin \left(\frac{\pi}{2}-x\right)+\right\}^3+\left\{\cos \left(\frac{\pi}{2}-x\right)\right\}^3 d x} d x \\ & \mathrm{I}=\int_0^{\pi / 2} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x \end{aligned}...(ii)  from  equation (i)+(ii)2I=∫0π/2sin⁡3x+cos⁡3xsin⁡3x+cos⁡3xdxor 2I=x=[π2−0]∴I=π4.\begin{aligned} & \text{\text{ from } equation }(i)+(i i) \\ & 2 \mathrm{I}=\int_0^{\pi / 2} \frac{\sin ^3 x+\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x \\ & \text{or } 2 \mathrm{I}=x=\left[\frac{\pi}{2}-0\right] \therefore \mathrm{I}=\frac{\pi}{4} .\end{aligned}
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