How do you find the integral of ${sin}^{2} (a x)$ $sin^2 (ax)$ ?

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Answer 1 · 2018-03-22T20:56:11

$\int {sin}^{2} (a x) d x = \frac{a x - sin (a x) cos (a x)}{2 a} + C$ $int sin^2(ax) dx= ( ax-sin(ax)cos(ax) )/(2a)+C$

Use the trigonometric identity:

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

So:

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

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

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

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

$\int {sin}^{2} (a x) d x = \frac{x}{2} - \frac{1}{2 a} sin (a x) cos (a x) + C$ $int sin^2(ax) dx= x/2 -1/(2a)sin(ax)cos(ax) +C$

$\int {sin}^{2} (a x) d x = \frac{a x - sin (a x) cos (a x)}{2 a} + C$ $int sin^2(ax) dx= ( ax-sin(ax)cos(ax) )/(2a)+C$

How do you find the integral of sin^2 (ax)sin2(ax) sin^2 (ax)?