What is $\int {sin}^{2} (x) d x$ $intsin^2(x) dx$ ?

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Answer 1 · 2014-12-15T15:03:02

We can solve this by expressing $\cdot {sin}^{2} (x)$ $*sin ^2 (x)$ in terms of a function in the first power.

From the equation:

$cos 2 x = {cos}^{2} (x) - {sin}^{2} (x)$ $cos 2x = cos ^ 2 (x) - sin^2 (x)$

and the equation

${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2 (x) + cos^2 (x) = 1$

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

then. we substitute to the first equation

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

simplifying,

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

substitute to the original problem

$\int [\frac{1 - cos 2 x}{2}] d x$ $int [(1-cos 2x)/2]dx$

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

$(\frac{1}{2}) [\int d x - \int cos 2 x d x$ $(1/2)[ intdx - intcos 2xdx$

$(\frac{1}{2}) [x - sin 2 x \cdot \int d (2 x)]$ $(1/2)[x - sin 2x*intd(2x)]$

$(\frac{1}{2}) [x - (sin 2 x) \cdot 2 \cdot 1]$ $(1/2)[x - (sin 2x) * 2*1]$

$\frac{x}{2} - sin 2 x + c$ $x/2 - sin 2x + c$

What is intsin^2(x) dx∫sin2(x)dxintsin^2(x) dx?