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

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

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Answer 1 · 2017-01-17T01:26:46

$\int x {sin}^{2} x d x = \frac{x^{2}}{4} - \frac{1}{4} x sin 2 x - \frac{1}{8} cos 2 x + C$ $int x sin^2x dx = x^2/4 -1/4 xsin2x -1/8cos2x + C$

Explanation:

Use the identity:

${sin}^{2} x = \frac{1 - cos 2 x}{2}$ $sin^2x = (1-cos2x)/2$

$\int x {sin}^{2} x d x = \int x (\frac{1 - cos 2 x}{2}) d x = \frac{1}{2} \int x d x - \frac{1}{2} \int x cos 2 x d x$ $int x sin^2x dx = int x((1-cos2x)/2 )dx = 1/2int xdx - 1/2int xcos2xdx$

The first integral is:

$\int x d x = \frac{x^{2}}{2} + C_{1}$ $int xdx = x^2/2+C_1$

the second can be integrated by part:

$\int x cos 2 x d x = \frac{1}{2} \int x d (sin 2 x) = \frac{1}{2} x sin 2 x - \frac{1}{2} \int sin 2 x d x = \frac{1}{2} x sin 2 x + \frac{1}{4} cos 2 x + C_{2}$ $int xcos2xdx = 1/2 int x d(sin2x) = 1/2xsin2x - 1/2 int sin2x dx = 1/2xsin2x +1/4 cos2x+C_2$

Putting it together:

$\int x {sin}^{2} x d x = \frac{x^{2}}{4} - \frac{1}{4} x sin 2 x - \frac{1}{8} cos 2 x + C$ $int x sin^2x dx = x^2/4 -1/4 xsin2x -1/8cos2x + C$

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

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