What is the antiderivative of ${cos}^{2} x$ $cos^2x$ ?

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What is the antiderivative of ${cos}^{2} x$ $cos^2x$ ?

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Answer 1 · 2015-06-08T00:16:42

The antiderivative is pretty much the same as the integral, except it's more general, so I'll do the indefinite integral.

$\int {cos}^{2} x d x$ $intcos^2xdx$

An identity for ${cos}^{2} x$ $cos^2x$ is:

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

$\Rightarrow \frac{1}{2} \int 1 + cos (2 x) d x$ $=> 1/2int 1+cos(2x)dx$

Since $\frac{d}{d x} [sin (2 x)] = 2 cos (2 x)$ $d/(dx)[sin(2x)] = 2cos(2x)$ , $\int cos (2 x) d x = \frac{1}{2} sin (2 x)$ $intcos(2x)dx = 1/2sin(2x)$ ;

$sin (2 x) = 2 sin x cos x$ $sin(2x) = 2sinxcosx$ , so $\frac{1}{2} sin (2 x) = sin x cos x$ $1/2sin(2x) = sinxcosx$

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

$= \frac{x}{2} + \frac{1}{4} sin (2 x) + C$ $= x/2 + 1/4sin(2x) + C$

$= \frac{x}{2} + \frac{1}{2} sin x cos x + C$ $= x/2 + 1/2sinxcosx + C$

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What is the antiderivative of ${cos}^{2} x$ $cos^2x$ ?

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