CBSE Class 12 Mathematics 2025: question 28(b)

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. Since $\sqrt{2\sin 2x}=\sqrt{4\sin x\cos x}=2\cos x\sqrt{\tan x}$, we get $I=\frac{1}{2}{\int }_0^{\pi /4}\frac{dx}{{\cos }^{4}x\sqrt{\tan x}}=\frac{1}{2}{\int }_0^{\pi /4}\frac{(1+{\tan }^{2}x){\sec }^{2}x}{\sqrt{\tan x}}dx$.

   ½ mark

2. Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$. When $x=0$, $t=0$; when $x=\frac{\pi }{4}$, $t=1$.

   ½ mark

3. Then $I=\frac{1}{2}{\int }_0^1\frac{1+t^2}{\sqrt{t}}dt=\frac{1}{2}{\int }_0^1\left(t^{-1/2}+t^{3/2}\right)dt$.

   ½ mark

4. Integrate: $I=\frac{1}{2}{\left[2\sqrt{t}+\frac{2}{5}{t}^{5/2}\right]}_0^1$.

   1 mark

5. Evaluate: $I=\frac{1}{2}\left(2+\frac{2}{5}\right)=\frac{1}{2}\times \frac{12}{5}=\frac{6}{5}$.

   ½ mark

Answer: $\frac{6}{5}$

Revise the chapter

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