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

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

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Answer 1 · 2017-03-15T17:49:05

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

Explanation:

Use the cosine double-angle identity to rewrite the function:

$cos (2 α) = 1 - 2 {sin}^{2} (α) \Rightarrow {sin}^{2} (α) = \frac{1 - cos (2 α)}{2}$ $cos(2alpha)=1-2sin^2(alpha)" "=>" "sin^2(alpha)=(1-cos(2alpha))/2$

Then:

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

So:

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

The second can be solved with the substitution $u = 4 x \Rightarrow d u = 4 d x$ $u=4x=>du=4dx$ :

$\int {sin}^{2} (2 x) d x = \frac{1}{2} x - \frac{1}{8} \int 4 cos (4 x) d x = \frac{1}{2} x - \frac{1}{8} \int cos (u) d u$ $intsin^2(2x)dx=1/2x-1/8int4cos(4x)dx=1/2x-1/8intcos(u)du$

The integral of cosine is sine:

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

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