Evaluate the following integrals: 2x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\text{x}\sin\text{x}\cos2\text{x dx}$

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Answer

$\int\text{x}.\cos2\text{x}\sin\text{x dx}$
$=\frac{1}{2}\int\text{x}(2\cos2\text{x}\sin\text{x})\text{dx}$ $\big[\therefore2\cos\text{A}\sin\text{B}=\sin(\text{A+B})-\sin(\text{A}-\text{B})\big]$
$=\frac{1}{2}\int\text{x}(\sin3\text{x}-\sin\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x}\sin3\text{x dx}-\frac{1}{2}\int\text{x}\sin\text{x dx}$
$=\frac{1}{2}\int\text{x}\sin3\text{x dx}-\frac{1}{2}\int\text{x}\sin\text{x dx}$
$=\frac{1}{2}\Big[\text{x}\int\sin3\text{x dx}-\int\Big\{\frac{\text{x}}{\text{dx}}(\text{x})\int\sin3\text{x dx}\Big\}\text{dx}\Big]\\-\frac{1}{2}\Big[\text{x}\int\sin\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x dx}\Big\}\text{dx}\Big]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)-\int1\Big(\frac{-\cos3\text{x}}{3}\Big)\text{dx}\Big]\\-\frac{1}{2}\big[\text{x}(-\cos\text{x})-\int1(-\cos\text{x})\text{dx}\big]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)+\frac{1}{9}\sin3\text{x}\big]-\frac{1}{2}\big[\text{x}(-\cos\text{x})+\sin\text{x}\big]$
$=-\frac{\text{x}\cos3\text{x}}{6}+\frac{\sin3\text{x}}{18}+\frac{\text{x}\cos\text{x}}{2}-\frac{\sin\text{x}}{2}+\text{C}$

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