What is $\int {cos}^{2} (4 x) d x$ $int cos^2(4x)dx$ ?

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Answer 1 · 2016-02-07T01:53:48

$\frac{1}{16} sin (8 x) + \frac{x}{2} + C$ $1/16sin(8x)+x/2+C$

In this instance we need to use a trig identity to re-express ${cos}^{2} (4 x)$ $cos^2(4x)$ in a form which can be integrated. We will use:

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

from the double angle formulae.

So re expressing the integral gives us:

$\int {cos}^{2} (4 x) d x = \int \frac{1}{2} cos (8 x) + \frac{1}{2} d x$ $int cos^2(4x)dx = int 1/2cos(8x)+1/2dx$

And now we can integrate easily to get:

$\int \frac{1}{2} cos (8 x) + \frac{1}{2} d x = \frac{1}{16} sin (8 x) + \frac{x}{2} + C$ $int 1/2cos(8x)+1/2dx = 1/16sin(8x)+x/2+C$

What is int cos^2(4x)dx ∫cos2(4x)dxint cos^2(4x)dx ?