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

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

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Answer 1 · 2015-11-03T15:30:41

$\int {sin}^{2} (x) d x = \frac{x}{2} - \frac{sin (2 x)}{4} + c$ $intsin^2(x)dx=x/2-sin(2x)/4+c$ ,
where $c$ $c$ is the constant of integration.

Explanation:

$\int {sin}^{2} (x) d x = \int \frac{d}{d x} (x) {sin}^{2} (x) d x$ $intsin^2(x)dx=intfrac{d}{dx}(x)sin^2(x)dx$

$= x {sin}^{2} (x) - \int x \frac{d}{d x} ({sin}^{2} (x)) d x$ $=xsin^2(x)-intxfrac{d}{dx}(sin^2(x))dx$

$= x {sin}^{2} (x) - \int x (2 sin (x) cos (x)) d x$ $=xsin^2(x)-intx(2sin(x)cos(x))dx$

$= x {sin}^{2} (x) - \int x sin (2 x) d x$ $=xsin^2(x)-intxsin(2x)dx$

$= x {sin}^{2} (x) + \frac{1}{2} \int x \frac{d}{d x} (cos (2 x)) d x$ $=xsin^2(x)+1/2intxfrac{d}{dx}(cos(2x))dx$

$= x {sin}^{2} (x) + \frac{1}{2} [x cos (2 x) - \int \frac{d}{d x} (x) cos (2 x) d x]$ $=xsin^2(x)+1/2[xcos(2x)-intfrac{d}{dx}(x)cos(2x)dx]$

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

$= x {sin}^{2} (x) + \frac{1}{2} x cos (2 x) - \frac{sin (2 x)}{4} + c$ $=xsin^2(x)+1/2xcos(2x)-sin(2x)/4+c$ ,
where $c$ $c$ is the constant of integration.

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

$= \frac{x}{2} - \frac{sin (2 x)}{4} + c$ $=x/2-sin(2x)/4+c$

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

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