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  • Please solve RD Sharma Class 12 Chapter Indefinite Integral Exercise Rivision Exercise Question 66 maths textbook solution.

Please solve RD Sharma Class 12 Chapter Indefinite Integral Exercise Rivision Exercise Question 66 maths textbook solution.

Answers (1)

Answer :

I=+\frac{1}{2} \ln |1+\cos x|-\frac{1}{10} \ln |1-\cos x|-\frac{3}{5} \ln |3 \cos x+2|+c

Hint: To solve the given solution we multiply and divide the given statement with sin x.

Given :  \int \frac{1}{\sin x(2+3 \cos x)} d x

Solution : 

\begin{aligned} &I=\int \frac{1}{\sin x(2+3 \cos x)} d x \\ &I=\int \frac{\sin x}{\sin ^{2} x(2+3 \cos x)} d x \\ &I=\int \frac{\sin x d x}{\left(1-\cos ^{2} x\right)(2+3 \cos x)} \end{aligned}

\begin{aligned} &t=\cos x, d t=-\sin x d x \\ &I=\int \frac{-d t}{\left(1-t^{2}\right)(2+3 t)} \\ &I=\int \frac{-d t}{(1+t)(1-t)(2+3 t)} \end{aligned}

\\\frac{1}{(1+t)(1-t)(2-3 t)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{c}{2+3 t}=A(1-t)(2+3 t)+B(1+t)(2+3 t)+C(1+t)(1-t)

\begin{array}{ll} A=\frac{1}{(1-(-1))(2-3)}=-\frac{1}{2} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because 1+t=0, t=-1]} \\ B=\frac{1}{(1+1)(2+3)}=\frac{1}{10} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because 1-t=0, t=1]} \\ C=\frac{1}{1-\left(\frac{4}{9}\right)}=\frac{9}{5} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\because 2+3 t=0, t=-\frac{2}{3}\right]} \end{array}

\begin{aligned} &I=-\int d t\left(\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{2+3 t}\right) \\ &I=-\left[A \int \frac{1}{1+t} d t+B \int \frac{1}{1-t} d t+C \int \frac{1}{2+3 t} d t\right. \end{aligned}

\begin{aligned} &I=-\left[-\frac{1}{2} \ln |1+t|+\frac{1}{10} \ln |1-t|+\frac{9}{5} \times \frac{1}{3} \ln |2+3 t|+c\right. \\ &I=+\frac{1}{2} \ln |1+t|-\frac{1}{10} \ln |1-t|-\frac{3}{5} \ln |3 t+2|+c \end{aligned}

I=+\frac{1}{2} \ln |1+\cos x|-\frac{1}{10} \ln |1-\cos x|-\frac{3}{5} \ln |3 \cos x+2|+c

Posted by

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