How do you integrate $int sin^2(x/5)*cos^3(x/5)$ ?

Question

1526 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-17T15:40:29

$5/3sin^3(x/5)-sin^5(x/5)+C$

We have

$intsin^2(x/5)cos^3(x/5)dx$

$intsin^2(x/5)cos^2(x/5)cos(x/5)dx$

Use $cos^2(x) = 1-sin^2(x)$ to re write the expression as:

$intsin^2(x/5)(1-sin^2(x/5))cos(x/5)dx$

$=int(sin^2(x/5)-sin^4(x/5))cos(x/5)dx$

Now apply the substitution:
$u = sin(x/5)$
$->du = 1/5cos(x)dx$

Which will give us the integral:

$5intu^2-u^4du$

$=5(u^3/3-u^5/5)+C$

Now reverse the substitution to get:

$5/3sin^3(x/5)-sin^5(x/5)+C$

How do you integrate int sin^2(x/5)*cos^3(x/5)int sin^2(x/5)*cos^3(x/5)?