Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 8

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 8

Answers (1)

best_answer

Answecgfy:  

\frac{1}{\sqrt{2}}(x)-\frac{1}{2 \sqrt{2}} \sin 2 x

Hint:

Use  \left[1-\cos x=2 \sin ^{2} x\right]

Given:

Let  I=\int \sin x \sqrt{1-\cos 2 x} d x
Solution:

I=\int \sin x \sqrt{1-\cos 2 x} d x

Using the above formula we get,

\begin{aligned} &\Rightarrow \sqrt{2} \int \sin x(\sin x) d x \\ &\Rightarrow \sqrt{2} \int \sin ^{2} x d x \Rightarrow \frac{+\sqrt{2}}{2} \int 1-\cos 2 x d x \\ &=\frac{1}{\sqrt{2}}(x)-\frac{1}{2 \sqrt{2}} \sin 2 x \end{aligned}

 

So the answer is   =\frac{1}{\sqrt{2}}(x)-\frac{1}{2 \sqrt{2}} \sin 2 x+c

 

 

Posted by

infoexpert27

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 practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question