# [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

Staff 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??
Your result is correct. W|A probably applied power reduction formulas to get their result. With trigonometric integrals, there are usually many ways to represent the result. If you are unsure of your result, differentiate it to ensure you get the original integrand.