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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
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Answer

Let $\text{I}=\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$ Then,$\text{I}=\int^\limits{\pi}_0\sin\text{x }\sin^2\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-2\cos^2\text{x})(1+\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-\cos\text{x})(1+2\cos\text{x})(1+\cos\text{x})^3\text{ dx}$
Let $\cos\text{x}=\text{t}$ Then, $-\sin\text{x dx}=\text{dt}$ When $\text{x}=0,\text{t}=1$ and $\text{x}=\pi,\text{t}=-1$$\therefore\ \text{I}=-\int^\limits{-1}_1(1-\text{t})(1+2\text{t})(1+\text{t})^3\text{ dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}-2\text{t}^2\big)\big(1+\text{t}^3+3\text{t}+3\text{t}^2\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}^3+3\text{t}+3\text{t}^3+\text{t}+\text{t}^4\\+3\text{t}^2+3\text{t}^3-2\text{t}^2-2\text{t}^5-6\text{t}^3-6\text{t}^4\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+4\text{t}+4\text{t}^2-2\text{t}^3-5\text{t}^4-2\text{t}^5\big)\text{dt}$
$\Rightarrow\text{I}=\Big[\text{t}+2\text{t}^2+\frac{4\text{t}^3}{3}-\frac{\text{t}^4}{2}-\text{t}^5-\frac{\text{t}^6}{3}\Big]^1_{-1}$
$\Rightarrow\text{I}=1+2+\frac{4}{3}-\frac{1}{2}-1-\frac{1}{3}\\+1-2+\frac{4}{3}+\frac{1}{2}-1+\frac{1}{3}$
$\Rightarrow\text{I}=\frac{8}{3}$

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