MHB Solve trig integral by substitution

[karush]

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

 

[MarkFL]

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

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

and then use the substitution:

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

and so we have:

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

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

 

[karush]

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??

 

[MarkFL]

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.