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  • 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

 

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