Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 38 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 38 Maths Textbook Solution.

Answers (1)

Answer: \frac{x^{3}}{3}\sin ^{-1}x+\frac{1}{3}\sqrt{1-x^{2}}-\frac{1}{9}\left ( 1-x^{2} \right )^{\frac{3}{2}}+c

Given:  \int x^{2}\sin ^{-1}xdx

Hint: \int uvdx=u\int vdx-\int \left ( \frac{d}{dx}u\int vdx \right )dx

Solution:

            I=\int x^{2}\sin ^{-1}x\: dx

          \begin{aligned} &I=\left(\sin ^{-1} x\right) \frac{x^{3}}{3}-\int \frac{1}{3 \sqrt{1-x^{2}}} \times x^{3} d x \ldots \text { applying byparts } \\ &I=\frac{x^{3}}{3} \sin ^{-1} x-\frac{1}{3} \int \frac{x^{2} x}{\sqrt{1-x^{2}}} d x \end{aligned}

                        Let 1-x^{2}=t => -2xdx = dt

            \begin{aligned} &=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{2} \times \frac{1}{3} \int \frac{(1-t) d t}{\sqrt{t}} \\ &=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{6}\left[\int \frac{1}{\sqrt{t}} d t-\int \frac{t}{\sqrt{t}} d t\right] \\ &=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{6}\left[2 \sqrt{t}-\frac{2 t^{\frac{3}{2}}}{3}\right] \\ &=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{3} \sqrt{1-x^{2}}-\frac{1}{9}\left(1-x^{2}\right)^{\frac{3}{2}}+c \end{aligned}

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question