Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 59 Maths Textbbok Solution.

Please Solve R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise  Revision Exercise Question 59 Maths Textbbok Solution.

Answers (1)

Answer:

I=-\frac{1}{2} \log |1+2 \cot x|+c

Given:

\int \frac{1}{\sin ^{2} x+\sin 2 x} d x

Hint:

To solve this equation we have to take sin²x common and also use sin2x formula.

Solution: 

I=\int \frac{1}{\sin ^{2} x+\sin 2 x} d x

    I=\int \frac{d x}{\sin ^{2} x+2 \sin x \cos x}

  I=\int \frac{d x}{\sin ^{2} x\left(1+\frac{2 \cos x}{\sin x}\right)}

I=\int \frac{\operatorname{cosec}^{2} x d x}{1+2 \cot x}

\text { Let } \cot x=t=>-\operatorname{cosec}^{2} x d x=d t

I=-\int \frac{d t}{1+2 t}

I=-\frac{1}{2} \log |1+2 t|+c \quad\left[\int \frac{d x}{x}=\log |x|+c\right]                            \left[\int \frac{d x}{a x+b}=\log \frac{(\mathrm{ax}+\mathrm{b})}{a}+c\right]

I=-\frac{1}{2} \log |1+2 \cot x|+c

 

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 reminds schools to submit registration data of Class 9, 11 students by October 16
anam-khan

CBSE reopens online portal for submission of LOC for Class 10, 12 board exams 2026; submit forms by Oct 8
anam-khan

CBSE private candidates challenge cancellation of post-Class 12 additional subject exam in Delhi HC

Latest Question