Get Answers to all your Questions

header-bg qa

Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 44.

Answers (1)

Answer  : \frac{\pi}{3}

Hint   : use indefinite formula and the limit to solve this integral

Given   : \int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x

Solution  :  \int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x

put x^{\frac{3}{2}}=t \Rightarrow \frac{3}{2} x^{\frac{3}{2}-1} d x=d t \Rightarrow \frac{3}{2} x^{\frac{1}{2}} d x=d t \Rightarrow d x=\frac{2}{3 \sqrt{x}} d t

when x=0 then t=0 when x=(\Pi)^{\frac{2}{3}} then t=\Pi



\int_{0}^{(\Pi)^{\frac{2}{3}}} \sqrt{x} \cos ^{2} x^{\frac{3}{2}} d x

\begin{aligned} &=\int_{0}^{\pi} \sqrt{x} \cos ^{2} t \frac{2}{3 \sqrt{x}} d t \\ &=\frac{2}{3} \int_{0}^{\pi} \cos ^{2} t d t \\ &=\frac{2}{3} \int_{0}^{\pi}\left(\frac{1+\cos 2 t}{2}\right) d t \\ &=\frac{1}{3} \int_{0}^{\pi}(1+\cos 2 t) d t \\ &=\frac{1}{3}\left[\int_{0}^{\pi} 1 d t\right]+\frac{1}{3}\left[\int_{0}^{\pi} \cos 2 t d t\right] \\ &=\frac{1}{3}[t]_{0}{ }^{\pi}+\frac{1}{3}\left[\frac{\sin 2 t}{2}\right]_{0}^{\pi} \\ &=\frac{1}{3} \Pi-0+\frac{1}{6}[\sin 2 \pi-\sin 2 \times 0] \end{aligned}





Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support