Evaluate the following integrals: ^3x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\text{x}\sin^3\text{x dx}$

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Answer

Let $\text{I}=\int\text{x}\sin^3\text{x dx}$
$\sin(3\text{A})=3\sin\text{A}-4\sin^3\text{A}$
$\sin^3\text{A}=\frac{1}4{}\big[3\sin\text{A}-\sin3\text{A}\big]$
$\therefore\text{I}=\frac{1}{4}\int\text{x}(3\sin\text{x}-\sin3\text{x})\text{dx}$
$=\frac{3}{4}\int\text{x}\sin\text{x dx}-\frac{1}{4}\int\text{x}\sin(3\text{x})\text{dx}$
$=\frac{3}{4}\big[\text{x}(-\cos\text{x})-\int1.(-\cos\text{x})\text{dx}\big]\\-\frac{1}{4}\Big[\text{x}\Big(-\frac{\cos3\text{x}}{3}\Big)-\int1.\Big(-\frac{\cos3\text{x}}{3}\Big)\text{dx}\Big]$
$=-\frac{3\text{x}\cos\text{x}}{4}+\frac{3}{4}\sin\text{x}+\frac{\text{x}\cos3\text{x}}{12}-\frac{1}{36}\sin3\text{x}+\text{C}$

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