Question

13.
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741 : (sin 3x.cos 6x dx.
Integrate : ſsin 3x.cos 6x dx .​

Answer 1

$\underline{\textsf{To find:}}$

$\mathsf{\displaystyle\int\,sin3x\,cos6x\;dx}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\textsf{We apply decomposition method to solve the given integral}$

$\mathsf{\displaystyle\int\,sin3x\,cos6x\;dx}$

$\textsf{This can be written as}$

$\mathsf{=\displaystyle\dfrac{1}{2}\int\,2\,sin3x\,cos6x\;dx}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{sin(A+B)+sin(A-B)=2\,sinA\,cosB}}$

$\mathsf{=\displaystyle\dfrac{1}{2}\int\,[sin(3x+6x)+sin(3x-6x)]\;dx}$

$\mathsf{=\displaystyle\dfrac{1}{2}\int\,[sin9x+sin(-3x)]\;dx}$

$\mathsf{=\displaystyle\dfrac{1}{2}\int\,[sin9x-sin3x]\;dx}$

$\mathsf{=\displaystyle\dfrac{1}{2}[\dfrac{-cos9x}{9}-(\dfrac{-cos3x}{3})]+C}$

$\mathsf{=\displaystyle\dfrac{1}{2}[\dfrac{-cos9x}{9}+\dfrac{cos3x}{3}]+C}$

$\underline{\textsf{Answer:}}$

$\mathsf{\displaystyle\int\,sin3x\,cos6x\;dx=\displaystyle\dfrac{1}{2}[\dfrac{-cos9x}{9}+\dfrac{cos3x}{3}]+C}$

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