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[SOLVED] solve trig integral by substitution

karush

Well-known member
Jan 31, 2012
2,727
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 31, 2012
2,727
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

Administrator
Staff member
Feb 24, 2012
13,775
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.