How do you integrate $\int cos 6 x d x$ $int cos6x dx$ ?

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Answer 1 · 2016-12-18T10:46:39

$\int cos (6 x) d x = \frac{1}{6} sin (6 x) + C$ $intcos(6x)dx=1/6sin(6x)+C$

Using integration by substitution along with the integral $\int cos (x) d x = sin (x) + C$ $intcos(x)dx = sin(x)+C$

let $u = 6 x \Rightarrow d u = 6 d x$ $u = 6x => du = 6dx$

then

$\int cos (6 x) d x = \frac{1}{6} \int cos (6 x) 6 d x$ $intcos(6x)dx = 1/6intcos(6x)6dx$

$= \frac{1}{6} \int cos (u) d u$ $=1/6intcos(u)du$

$= \frac{1}{6} sin (u) + C$ $=1/6sin(u)+C$

$= \frac{1}{6} sin (6 x) + C$ $=1/6sin(6x)+C$

How do you integrate ∫cos6xdxint cos6x dx?