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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Definite Integration

NDA 2019 · Mathematics · Definite Integration

Paragraph: Consider the integrals I_1=∫_0^x (x d x)/(1+ x) and I...

Paragraph: Consider the integrals I1=∫0xxdx1+sin⁡x and \mathrm{I}_1=\int_0^x \frac{x d x}{1+\sin x} \text{ and }\mathrm{I}2=∫0π(π−x)dx1−sin⁡(π+x)_2=\int_0^\pi \frac{(\pi-x) d x}{1-\sin (\pi+x)}Question: What is the value of I1I_1 ?

Options

  1. A0
  2. Bπ2\frac{\pi}{2}
  3. Cπ\pi
  4. D2π2 \pi

Correct answer

C. π\pi

Step-by-step solution

I1=∫0πxdx1+sin⁡x=∫0ππ−x1+sin⁡(π−x)dx⇒I1=∫0ππ−x1+sin⁡x⋅dx⇒21=∫0ππ1+sin⁡x⋅dx⇒I1=π2∫0π11+cos⁡(π2−x)⋅dx⇒I1=π2∫0π11+2cos⁡2(π4−x2)−1⋅dx\begin{aligned} & 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}⇒I1=π2∫0πsec⁡2(π4−x2)⋅dx⇒I1=π2[tan⁡(π4−x2)−1]0π⇒I1=−π2(−1−1)=π\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}$
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