How do you find the antiderivative of ${sin}^{4} (x)$ $sin^4(x)$ ?

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How do you find the antiderivative of ${sin}^{4} (x)$ $sin^4(x)$ ?

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Answer 1 · 2015-04-29T18:51:59

You could use power reduction formulas (double angle)

Because $cos 2 θ = 1 - 2 {sin}^{2} θ = 2 cos θ - 1$ $cos 2 theta = 1-2sin^2 theta = 2cos theta -1$ , we get

${sin}^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ $sin^2theta = 1/2 (1-cos2 theta)$
and
${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2theta = 1/2 (1+cos2 theta)$

$\int {sin}^{4} x d x = \int {({sin}^{2} x)}^{2} d x$ $int sin^4x dx = int (sin^2x)^2 dx$

$sssssssssss$ $color(white)"sssssssssss"$ $= \int {[\frac{1}{2} (1 - cos 2 x)]}^{2} d x$ $= int [1/2(1-cos2x)]^2 dx$

$sssssssssss$ $color(white)"sssssssssss"$ $= \frac{1}{4} \int (1 - 2 cos 2 x + {cos}^{2} 2 x) d x$ $= 1/4 int (1-2cos2x+ cos^2 2x) dx$

$sssssssssss$ $color(white)"sssssssssss"$ $= \frac{1}{4} \int (1 - 2 cos 2 x + \frac{1}{2} (1 + cos 4 x)) d x$ $= 1/4 int (1-2cos2x+ 1/2(1+cos4x)) dx$

$sssssssssss$ $color(white)"sssssssssss"$ $= \frac{1}{4} \int (\frac{3}{2} - 2 cos 2 x + \frac{1}{2} cos 4 x) d x$ $= 1/4 int (3/2-2cos2x+ 1/2 cos4x) dx$

$sssssssssss$ $color(white)"sssssssssss"$ $= \frac{1}{4} (\frac{3}{2} x - sin 2 x + \frac{1}{8} sin 4 x) + C$ $= 1/4 (3/2 x - sin2x+ 1/8 sin4x) +C$

$sssssssssss$ $color(white)"sssssssssss"$ $= \frac{3}{8} x - \frac{1}{4} sin 2 x + \frac{1}{32} sin 4 x + C$ $= 3/8 x- 1/4 sin2x+ 1/32 sin4x +C$

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How do you find the antiderivative of ${sin}^{4} (x)$ $sin^4(x)$ ?

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How do you find the antiderivative of ${sin}^{4} (x)$ $sin^4(x)$ ?