CBSE Class 12 Mathematics 2023: question 33

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. $I={\int }_0^{\pi /2}\sin 2x{\tan }^{-1}(\sin x)dx={\int }_0^{\pi /2}2\sin x\cos x{\tan }^{-1}(\sin x)dx$

   ½ mark

2. Put $\sin x=t$, so $\cos xdx=dt$.

   When $x=0$, $t=0$; when $x=\frac{\pi }{2}$, $t=1$.

   ½ mark

3. $I=2{\int }_0^{1}t{\tan }^{-1}tdt$

   ½ mark

4. Integrating by parts, with ${\tan }^{-1}t$ as the first function:

   $I=2\left[\frac{t^2}{2}{\tan }^{-1}t{|}_0^1-{\int }_0^1\frac{1}{1+t^2}\cdot \frac{t^2}{2}dt\right]$

   1 mark

5. $=2\cdot \frac{1}{2}\cdot \frac{\pi }{4}-{\int }_0^1\frac{t^2}{1+t^2}dt=\frac{\pi }{4}-{\int }_0^1\left[1-\frac{1}{1+t^2}\right]dt$

   1 mark

6. $=\frac{\pi }{4}-[t{]}_0^1+[{\tan }^{-1}t{]}_0^1=\frac{\pi }{4}-1+\frac{\pi }{4}$

   1 mark

7. $I=\frac{\pi }{2}-1$

   ½ mark

Answer: $\frac{\pi }{2}-1$

Revise the chapter

• Chapter: Integrals (notes, formulas, where marks are lost and every past question)