1. ## Calculus - Integration

how would I go about integrating (cos x)^3?

2. change

$\cos^3{x}= \cos^2{x}\cos{x}$

3. Originally Posted by 11rdc11
change

$\cos^3{x}= \cos^2{x}\cos{x}$
Alternatively

Since $\cos{3x} = 4\cos^3{x} - 3\cos{x}$

$4\cos^3{x} = \cos{3x} + 3\cos{x}$

$\cos^3{x} = \frac{1}{4}\cos{3x} + \frac{3}{4}\cos{x}$.

Therefore

$\int{\cos^3{x}\,dx} = \int{\frac{1}{4}\cos{3x} + \frac{3}{4}\cos{x}\,dx}$

$= \frac{1}{12}\sin{3x} + \frac{3}{4}\sin{x} + C$.

4. Originally Posted by 11rdc11
change

$\cos^3{x}= \cos^2{x}\cos{x}$
@OP: Then replace $\cos^2 x$ with $1 - \sin^2 x$. Then make the substitution $u = \sin x$.