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

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

Answers (1)

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}

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