What is the integral of $sin (x) cos (x)$ $sin(x) cos(x)$ ?

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What is the integral of $sin (x) cos (x)$ $sin(x) cos(x)$ ?

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Answer 1 · 2014-12-16T05:02:06

It's $\frac{1}{2} {sin}^{2} (x) + C$ $1/2 sin^2 (x) + C$ .

The substitution used to solve this integral is simple.

Note that $cos (x)$ $cos(x)$ is the derivative of $sin (x)$ $sin(x)$ .

Define the variable $u = sin (x)$ $u=sin(x)$ .
We have $d u = cos (x) d x$ $du=cos(x)dx$ .
So, $d x = \frac{1}{cos (x)} d u$ $dx=1/cos(x) du$ .

The integral:

$\int sin (x) cos (x) d x = \int u \frac{cos (x)}{cos (x)} d u = \int u d u$ $int sin(x) cos(x) dx = int u cos(x)/cos(x) du = int u du$

Knowing that $\frac{d}{d u} [\frac{1}{2} u^{2} + C] = u$ $d/(du)[1/2 u^2 + C]=u$ we have:

$\int u d u = \int \frac{d}{d u} [\frac{1}{2} u^{2} + C] d u$ $int u du = int d/(du)[1/2 u^2 + C] du$

Using the Fundamental Theorem of Calculus we get:

$\int \frac{d}{d u} [\frac{1}{2} u^{2} + C] d u = \frac{1}{2} u^{2} + C = \frac{1}{2} {sin}^{2} (x) + C$ $int d/(du)[1/2 u^2 + C] du = 1/2 u^2 + C = 1/2 sin^2 (x) + C$

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