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 12th result 2025 sees more in compartment, JNVs on top, perfect-scorers in minor subjects
anam-khan

Low score in CBSE results? Courses, career options after 12th open to you
anam-khan

CBSE class 10, 12 results: West Bengal CM Mamata Banerjee congratulates successful candidates

Latest Question