How do you evaluate the integral of $\int {(1 + cos 4 x)}^{\frac{3}{2}} d x$ $int (1 + cos 4x)^(3/2) dx$ ?

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How do you evaluate the integral of $\int {(1 + cos 4 x)}^{\frac{3}{2}} d x$ $int (1 + cos 4x)^(3/2) dx$ ?

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Answer 1 · 2018-03-28T21:50:27

$\int {(1 + cos 4 x)}^{\frac{3}{2}} d x = \frac{\sqrt{2}}{8} (3 sin 2 x + sin 6 x) + c$ $int(1+cos4x)^(3/2)dx=sqrt2/8(3sin2x+sin6x)+"c"$

Explanation:

$\int {(1 + cos 4 x)}^{\frac{3}{2}} d x = \int {(1 + 2 {cos}^{2} 2 x - 1)}^{\frac{3}{2}} d x = \int 2 \sqrt{2} {cos}^{3} 2 x d x = \frac{\sqrt{2}}{2} \int {(2 {cos}^{2} x - 1)}^{3} d x$ $int(1+cos4x)^(3/2)dx=int(1+2cos^2 2x-1)^(3/2)dx=int2sqrt2cos^3 2xdx=sqrt2/2int(2cos^2x-1)^3dx$

After a very long expansion, we get it into the following form

$\frac{\sqrt{2}}{2} \int 3 cos 2 x + cos 6 x d x = \frac{\sqrt{2}}{2} (\frac{1}{2} sin 2 x + \frac{1}{6} sin 6 x) + c$ $(sqrt2)/2int3cos2x+cos6xdx=sqrt2/2(1/2sin2x+1/6sin6x)+"c"$
$= \frac{\sqrt{2}}{8} (3 sin 2 x + sin 6 x) + c$ $=sqrt2/8(3sin2x+sin6x)+"c"$

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How do you evaluate the integral of $\int {(1 + cos 4 x)}^{\frac{3}{2}} d x$ $int (1 + cos 4x)^(3/2) dx$ ?

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Explanation:

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How do you evaluate the integral of $\int {(1 + cos 4 x)}^{\frac{3}{2}} d x$ $int (1 + cos 4x)^(3/2) dx$ ?