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Question

Integrate:
$$\int \sin^4xdx$$


Solution

$$ \displaystyle  I = \int sin^4 x dx = \int \frac{(1-cos 2x)^2}{4}dx $$

$$ [\because \dfrac{1-cos2x}{2} = sinx]$$

$$ \displaystyle  = \frac{1}{4} \int (1-2cos2x+cos^2 2x )dx $$

$$ \displaystyle =\frac{1}{4} \int 1-2cos2x+\frac{(1+cos4x)}{2}dx$$

$$ \displaystyle =  \frac{1}{4}\int (\frac{3}{2}-2cos2x+\frac{cos4x}{2})dx$$

$$ \displaystyle = \frac{1}{4} (\frac{3}{2}x - \frac{2sin2x}{2}+\frac{sin{4}x}{2})+c$$

$$ I = \dfrac{3}{8}x-\dfrac{1}{4}sin2x + \dfrac{1}{32}sin4x+c $$

Mathematics

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