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  • Please solve RD Sharma Class 12 Chapter Indefinite Integral Exercise Rivision Exercise Question 67 maths textbook solution.

Please solve RD Sharma Class 12 Chapter Indefinite Integral Exercise Rivision Exercise Question 67 maths textbook solution.

Answers (1)

Answer :

=\frac{1}{6} \ln (1-\cos x)+\frac{1}{2} \ln (1+\cos x)-\frac{2}{3} \ln (1+2 \cos x)+c

Hint  : To solve the given statement we will use partial fraction rule.

Given :  \int \frac{1}{\sin x+\sin 2 x} d x

Solution :

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

\\\frac{1}{(1-t)(1+t)(1+2 t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{c}{1+2 t}=A(1+t)(1+2 t)+B(1-t)(1+2 t)+C(1-t)(1+t)

\begin{aligned} &1=A+2 A t+A t+2 A t^{2}+B+2 B t-B t-2 B t^{2}+C-C t^{2} \\ &0=2 A-2 B-C \ldots \ldots \ldots . . (1)\end{aligned}

\begin{aligned} &0=3 A+B \ldots \ldots \ldots \ldots \text { (2) } \quad B=-3 A \\ &1=A+B+C \ldots \ldots \ldots \text { (3) } \end{aligned}

On solving, A=\frac{1}{6}, B=-\frac{1}{2}, C=\frac{4}{3}

\int \frac{1}{(1-t)(1+t)(1+2 t)} d t=-\int \frac{1}{6(1-t)}+\int \frac{1}{2(1+t)}-\int \frac{4}{3} \frac{1}{(1+2 t)} d t

\begin{aligned} &=-\frac{1}{6}-\ln (1-t)+\frac{1}{2} \ln (1+t)-\frac{4}{3} \times \frac{1}{2} \ln (1+2 t) \\ &=\frac{1}{6} \ln (1-\cos x)+\frac{1}{2} \ln (1+\cos x)-\frac{2}{3} \ln (1+2 \cos x)+c \end{aligned}

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