How do you integrate $\int {sin}^{4} (2 x)$ $int sin^(4) (2x)$ ?

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How do you integrate $\int {sin}^{4} (2 x)$ $int sin^(4) (2x)$ ?

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Answer 1 · 2017-01-29T22:10:12

$\int {sin}^{4} (2 x) d x = \frac{3}{8} x - \frac{1}{8} sin (4 x) + \frac{1}{64} sin (8 x) + C$ $intsin^4(2x)dx=3/8x-1/8sin(4x)+1/64sin(8x)+C$

Explanation:

$I = \int {sin}^{4} (2 x) d x$ $I=intsin^4(2x)color(red)(dx)$

A great way to deal with even powers of sine and cosine, assuming a substitution won't work, is to use the cosine double-angle identity.

The form of the cosine double-angle identity involving sine is $cos (2 θ) = 1 - 2 {sin}^{2} (θ)$ $cos(2theta)=1-2sin^2(theta)$ . This can be solved for ${sin}^{2} (θ)$ $sin^2(theta)$ as ${sin}^{2} (θ) = \frac{1 - cos (2 θ)}{2}$ $sin^2(theta)=(1-cos(2theta))/2$ .

Notice that the following are analogous forms:

${sin}^{2} (θ) = \frac{1 - cos (2 θ)}{2} \Rightarrow {sin}^{2} (2 x) = \frac{1 - cos (4 x)}{2}$ $sin^2(theta)=(1-cos(2theta))/2" "=>" "color(blue)(sin^2(2x)=(1-cos(4x))/2$

Whatever is in the argument of the cosine function must be double that in the sine function. Using the blue identity, we can then rewrite the integrand:

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

From here, expand ${(1 - cos (4 x))}^{2}$ $(1-cos(4x))^2$ :

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

$I = \frac{1}{4} \int d x - \frac{1}{2} \int cos (4 x) d x + \frac{1}{4} \int {cos}^{2} (4 x) d x$ $I=1/4intdx-1/2intcos(4x)dx+1/4intcos^2(4x)dx$

The first two integrals are solved fairly easily. The second one can be solved using the chain rule in reverse (with the substitution $u = 4 x \Rightarrow d u = 4 . d x$ $u=4x=>du=4color(white).dx$ ).

$I = \frac{1}{4} x - \frac{1}{8} \int 4 cos (4 x) d x + \frac{1}{4} \int {cos}^{2} (4 x) d x$ $I=1/4x-1/8int4cos(4x)dx+1/4intcos^2(4x)dx$

$I = \frac{1}{4} x - \frac{1}{8} sin (4 x) + \frac{1}{4} \int {cos}^{2} (4 x) d x$ $I=1/4x-1/8sin(4x)+1/4intcos^2(4x)dx$

To resolve this even power of cosine, use another form of the cosine double-angle identity: $cos (2 θ) = 2 {cos}^{2} (θ) - 1$ $cos(2theta)=2cos^2(theta)-1$ . This shows that ${cos}^{2} (θ) = \frac{1 + cos (2 θ)}{2}$ $cos^2(theta)=(1+cos(2theta))/2$ .

Which is analogous to:

${cos}^{2} (θ) = \frac{1 + cos (2 θ)}{2} \Rightarrow {cos}^{2} (4 x) = \frac{1 + cos (8 x)}{2}$ $cos^2(theta)=(1+cos(2theta))/2" "=>" "color(green)(cos^2(4x)=(1+cos(8x))/2)$

Substituting this in:

$I = \frac{1}{4} x - \frac{1}{8} sin (4 x) + \frac{1}{4} \int \frac{1 + cos (8 x)}{2} d x$ $I=1/4x-1/8sin(4x)+1/4int(1+cos(8x))/2dx$

Which can then be split up and integrated as before:

$I = \frac{1}{4} x - \frac{1}{8} sin (4 x) + \frac{1}{8} \int d x + \frac{1}{8} \int cos (8 x) d x$ $I=1/4x-1/8sin(4x)+1/8intdx+1/8intcos(8x)dx$

$I = \frac{1}{4} x - \frac{1}{8} sin (4 x) + \frac{1}{8} x + \frac{1}{64} \int 8 cos (8 x) d x$ $I=1/4x-1/8sin(4x)+1/8x+1/64int8cos(8x)dx$

$I = \frac{3}{8} x - \frac{1}{8} sin (4 x) + \frac{1}{64} sin (8 x) + C$ $I=3/8x-1/8sin(4x)+1/64sin(8x)+C$

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