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

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

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Answer 1 · 2016-11-24T13:21:52

Use trigonometry to rewrite.

Explanation:

When studying trigonometry, some encounter formulas called "Power Reduction Formulas". These are not used for anything -- until one studies integration. Many (most?) students who encounter these formulas forget them before they need them.

$cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B) = cosAcosB-sinAsinB$ , so

$cos (2 A) = {cos}^{2} A - {sin}^{2} A$ $cos(2A) = cos^2A - sin^2A$ and since ${cos}^{2} A = 1 - {sin}^{2} A$ $cos^2A = 1-sin^2A$ ,

$cos (2 A) - 1 - 2 {sin}^{2} A$ $cos(2A) - 1-2sin^2A$ .

If we solve for $sin (A)$ $sin(A)$ , we have the half angle formula, but is we stop at solving for ${sin}^{2} (A)$ $sin^2(A)$ , the we have the power reduction formula for $sin x$ $sinx$

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

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

$= \frac{1}{2} [x - \frac{1}{2} sin (2 x)] + C$ $= 1/2[x-1/2sin(2x)] +C$ .

Rewrite the answer to taste.

I like $\frac{1}{2} (x - sin x cos x) + C .$ $1/2(x-sinxcosx) +C.$

Additional note

Using the cosine double angle formula in the form

$cos (2 A) = 2 {cos}^{2} A - 1$ $cos(2A) = 2cos^2A - 1$ , we get the power reduction for ${cos}^{2} x$ $cos^2x$

${cos}^{2} x = \frac{1}{2} (1 + cos (2 x))$ $cos^2x = 1/2(1+cos(2x))$

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