Q244 / 270 · NDA 2019 (Phase 2) · Mathematics

Paragraph: Consider the integrals $\mathrm{I}_1=\int_0^x \frac{x d x}{1+\sin x} \text { and } \mathrm{I}_2=\int_0^\pi \frac{(\pi-x) d x}{1-\sin (\pi+x)}$
Question: What is the value of $I_1$ ?

  • A
    0
  • B
    $\frac{\pi}{2}$
  • Correct answer
    C
    $\pi$
  • D
    $2 \pi$

Solution

$\begin{aligned} & \text { } I_1=\int_0^\pi \frac{x d x}{1+\sin x}=\int_0^\pi \frac{\pi-x}{1+\sin (\pi-x)} d x \\ & \Rightarrow I_1=\int_0^\pi \frac{\pi-x}{1+\sin x} \cdot d x \\ & \Rightarrow 2_1=\int_0^\pi \frac{\pi}{1+\sin x} \cdot d x \\ & \Rightarrow I_1=\frac{\pi}{2} \int_0^\pi \frac{1}{1+\cos \left(\frac{\pi}{2}-x\right)} \cdot d x \\ & \Rightarrow I_1=\frac{\pi}{2} \int_0^\pi \frac{1}{1+2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)-1} \cdot d x\end{aligned}$ $\begin{aligned} & \Rightarrow I_1=\frac{\pi}{2} \int_0^\pi \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot d x \\ & \Rightarrow I_1=\frac{\pi}{2}\left[\frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)}{-1}\right]_0^\pi \\ & \Rightarrow I_1=\frac{-\pi}{2}(-1-1)=\pi\end{aligned}$

Hence, the correct answer is option: 'C'.

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