What is the integral of cos^3(x)cos(3x) ?

The integral of cos^3(x)cos(3x) is

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integral of cos^3(x)cos(3x)

\int cos^{3} (x) cos (3 x) d x

\frac{1}{2} cos^{2} (x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) + \frac{1}{2} (\frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} (x - \frac{1}{4} sin (4 x))) + C

Solution steps

\int cos^{3} (x) cos (3 x) d x

Simplify

cos^{3} (x) : cos^{2} (x) cos (x)

= \int cos^{2} (x) cos (x) cos (3 x) d x

Apply Integration By Parts

: \frac{1}{2} cos^{2} (x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) - \int - \frac{1}{2} sin (2 x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) d x

= \frac{1}{2} cos^{2} (x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) - \int - \frac{1}{2} sin (2 x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) d x

\int - \frac{1}{2} sin (2 x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) d x = - \frac{1}{2} (\frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} (x - \frac{1}{4} sin (4 x)))

= \frac{1}{2} cos^{2} (x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) - (- \frac{1}{2} (\frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} (x - \frac{1}{4} sin (4 x))))

Simplify

= \frac{1}{2} cos^{2} (x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) + \frac{1}{2} (\frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} (x - \frac{1}{4} sin (4 x)))

Add a constant to the solution

= \frac{1}{2} cos^{2} (x) (\frac{1}{4} sin (4 x) + \frac{1}{2} sin (2 x)) + \frac{1}{2} (\frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} (x - \frac{1}{4} sin (4 x))) + C

n \to \infty lim (\frac{1}{n ^{2} + 2 n - n})

\frac{sec ( x )}{7 + sec ( x )}

y^{^{''}} - 36 y^{^{'}} = 0

\int - 2 e^{- x} x d x

\int \frac{1}{x ^{2} + 1} d x