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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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