Integral of cos^4(x/4) dx ?

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Integral of cos^4(x/4) dx ?

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Answer 1 · 2018-02-26T17:24:17

$I = \frac{1}{8} (3 x + 8 sin (\frac{x}{2}) + sin (x))$ $I=1/8(3x+8sin(x/2)+sin(x))$

Explanation:

We want to solve

$I = \int {cos}^{4} (\frac{x}{4}) d x$ $I=intcos^4(x/4)dx$

When we have an integral of a trigonmetric function,
to some power, we often want to reduce the power of the trigonometric function, which will make the integration easier

So let's reduce the power of cosine, using

${cos}^{2} (x) = \frac{1}{2} (1 + cos (2 x))$ $cos^2(x)=1/2(1+cos(2x))$

Thus

${cos}^{4} (\frac{x}{4}) = {({cos}^{2} (\frac{x}{4}))}^{2}$ $cos^4(x/4)=(cos^2(x/4))^2$

$= {(\frac{1}{2} (1 + cos (\frac{x}{2})))}^{2}$ $=(1/2(1+cos(x/2)))^2$

$= \frac{1}{4} {(1 + cos (\frac{x}{2}))}^{2}$ $=1/4(1+cos(x/2))^2$

$= \frac{1}{4} (1 + 2 cos (\frac{x}{2}) + {cos}^{2} (\frac{x}{2}))$ $=1/4(1+2cos(x/2)+cos^2(x/2))$

$= \frac{1}{4} (1 + 2 cos (\frac{x}{2}) + \frac{1}{2} (1 + cos (x)))$ $=1/4(1+2cos(x/2)+1/2(1+cos(x)))$

$= \frac{1}{4} (\frac{3}{2} + 2 cos (\frac{x}{2}) + \frac{1}{2} cos (x))$ $=1/4(3/2+2cos(x/2)+1/2cos(x))$

So the integral becomes

$I = \frac{1}{4} \int (\frac{3}{2} + 2 cos (\frac{x}{2}) + \frac{1}{2} cos (x)) d x$ $I=1/4int(3/2+2cos(x/2)+1/2cos(x))dx$

$= \frac{1}{4} (\frac{3}{2} x + 4 sin (\frac{x}{2}) + \frac{1}{2} sin (x)) + C$ $=1/4(3/2x+4sin(x/2)+1/2sin(x))+C$

$= \frac{1}{8} (3 x + 8 sin (\frac{x}{2}) + sin (x)) + C$ $=1/8(3x+8sin(x/2)+sin(x))+C$

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Integral of cos^4(x/4) dx ?

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