Notes: How to find out the indefinite integral of the cube of sinx?

Friday, August 16, 2013

$\displaystyle\int$ (sinx)^3dx = $\displaystyle\int$ [(sinx)^2*sin
x]dx

According to Pythagorean identity, we'll
get:

(sinx)^2 = 1 -
(cosx)^2

$\displaystyle\int$ [(sinx)^2*sin x]dx = $\displaystyle\int$ [(1 -
(cosx)^2)*sin x]dx

We'll remove the
brackets:

$\displaystyle\int$ [(1 - (cosx)^2)*sin x]dx = $\displaystyle\int$ sin xdx -
$\displaystyle\int$ (cosx)^2*sin xdx

We'll evaluate $\displaystyle\int$ (cosx)^2*sin
xdx using substitution:

cos x =
t

We'll differentiate both
sides:

-sin x dx = dt

We'll
re-write the integral, changing the variable:

$\displaystyle\int$
(cosx)^2*sin xdx = - $\displaystyle\int$ t^2dt

- $\displaystyle\int$ t^2dt = -t^3/3 +
C

$\displaystyle\int$ (cosx)^2*sin xdx = -(cos x)^3/3 +
C

$\displaystyle\int$ (sinx)^3dx = $\displaystyle\int$ sin xdx - $\displaystyle\int$ (cosx)^2*sin
xdx

$\displaystyle\int$ (sinx)^3dx = -cos x + (cos x)^3/3 +
C

Notes