# [SOLVED]solve trig integral by substitution

#### karush

##### Well-known member
this is supposed to be solved with U substitution
$\displaystyle \int \sin^6(x)\cos^3(x)$
since $$\displaystyle \cos^3(x)$$ has an odd power then
$\displaystyle \int \sin^6(x)\left(1-\sin^2(x)\right)\cos(x) dx$
then substitute $$\displaystyle u=\sin(x)$$ and $$\displaystyle du=cos(x)dx$$
$\int u^6 \left(1-u^2\right) du$
so if ok so far

#### MarkFL

Staff member
Yes, that looks good. Like you, I would seek an integral of the form:

$$\displaystyle \int f\left(\sin(x) \right)\,\cos(x)\,dx$$

and then use the substitution:

$$\displaystyle u=\sin(x)\,\therefore\,du=\cos(x)\,dx$$

and so we have:

$$\displaystyle \int f(u)\,du$$

So, now just integrate with respect to $u$, then back substitute for $u$ to express the anti-derivative in terms of $x$.

#### karush

##### Well-known member
so from
$\int u^6 \left(1-u^2\right) du$
by distribution

$\int \left(u^6-u^8\right) du$

$\left(\frac{u^7}{7}-\frac{u^9}{9}\right)$

$\frac{sin^7(x)}{7}-\frac{sin^9(x)}{9}+C$

W|A gave a much more complicated answer than this??

Last edited:

#### MarkFL

$\int u^6 \left(1-u^2\right) du$
$\int \left(u^6-u^8\right) du$
$\left(\frac{u^7}{7}-\frac{u^9}{9}\right)$
$\frac{sin^7(x)}{7}-\frac{sin^9(x)}{9}+C$