#### Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 48.

Answer :  $\frac{8}{3}$

Hint:  Use indefinite formula and the given limits to solve this integral

Given:

$\int_{0}^{\pi}\sin^3x(1+2\cos x)(1+\cos x)^2dx$

Solution:

$\int_{0}^{\pi}\sin^3x(1+2\cos x)(1+\cos x)^2dx$

put $(\cos x)=t$

$\Rightarrow -\sin x\; dx=dt$

$\Rightarrow dx=dt - \sin x$
When x=0  then t=1

when $x=\pi$  then t=-1

\begin{aligned} &\therefore \int_{0}^{\pi} \sin ^{3} x(1+2 \cos x)(1+\cos x)^{2} d x \\ &=\int_{1}^{-1} \sin ^{3} x(1+2 t)(1+t)^{2} \frac{d t}{-\sin x} \\ &=-\int_{+1}^{-1} \sin ^{2} x(1+2 t)\left(1+t^{2}+2 t\right) d t \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 \alpha b\right] \end{aligned}

\begin{aligned} &=-\int_{+1}^{-1}\left(1-\cos ^{2} x\right)\left(1+t^{2}+2 t+2 t+2 t^{3}+4 t^{2}\right) d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because 1=\sin ^{2} \theta+\cos ^{2} \theta\right] \\ &=-\int_{+1}^{-1}\left(1-t^{2}\right)\left(1+5 t^{2}+4 t+2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(t^{2}-1\right)\left(1+5 t^{2}+4 t+2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(t^{2}+5 t^{4}+4 t^{3}+2 t^{5}-1-5 t^{2}-4 t-2 t^{3}\right) d t \\ &=\int_{+1}^{-1}\left(2 t^{5}+2 t^{3}+5 t^{4}-4 t^{2}-4 t-1\right) d t \end{aligned}

\begin{aligned} &=2 \int_{1}^{-1} t^{5} d t+2 \int_{1}^{-1} t^{3} d t+5 \int_{1}^{-1} t^{4} d t-4 \int_{.1}^{-1} t^{2} d t-4 \int_{-1}^{-1} t d t-\int_{1}^{-1} d t \\ &=2\left[\left.\frac{t^{5+1}}{5+1}\right|_{1} ^{-1}+2\left[\frac{t^{3+1}}{3+1}\right]_{1}^{-1}+5\left[\left.\frac{t^{4+1}}{4+1}\right|_{1} ^{-1}-4\left[\left.\frac{t^{2+1}}{2+1}\right|_{1} ^{-1}-4\left[\frac{t^{1+1}}{1+1}\right]_{1}^{-1}-\left[\left.t\right|_{1} ^{-1}\right.\right.\right.\right. \\ &=\frac{2}{6}\left[t^{6}\right]_{1}^{-1}+\frac{2}{4}\left[t^{4}\right]_{1}^{-1}+\frac{5}{5}\left[t^{5}\right]_{1}^{-1}-\frac{4}{3}\left[t^{3}\right]_{1}^{-1}-4\left[\frac{t^{2}}{2}\right]_{1}^{-1}-[t]_{1}^{-1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \end{aligned}

\begin{aligned} &=\frac{2}{6}\left[(-1)^{6}-1^{6}\right]+\frac{2}{4}\left[(-1)^{4}-1^{4}\right]+\left[(-1)^{5}-1^{5}\right]-\frac{4}{3}\left[(-1)^{3}-1\right]-\frac{4}{2}\left[(-1)^{2}-1\right]-[-1-1] \\ &=\frac{1}{3}[1-1]+\frac{1}{2}[1-1]+[-1-1]-\frac{4}{3}[-1-1]-2[1-1]-(-2) \\ &=0+0+(-2)-\frac{4}{3}(-2)-2 \times 0+2 \\ &=\frac{8}{3} \end{aligned}