1. ## Trig Antiderivative

$\int sec(x)sin^3(x)dx$

$\int \frac {tan(x)}{sin(x)} sin^3(x)dx$

So then the sin(x)'s cancel out leaving...
[tex] \int tan(x)sin^2(x)dx

then

$\int \frac {sin(x)}{cos(x)} sin^2(x)dx$

this equals

$\int \frac {sin^3(x)}{cos(x)}dx$

Then split up the $sin^3(x)$ and then set the $sin^2(x)$ equal to $1 - cos^2(x)$

$\int \frac {1 - cos^2(x)}{cos(x)} sin(x)dx$

Then the cos(x)'s can cancel...

$\int (1 - cos(x)) sin(x)dx$

this then equals

$\int sin(x) - cos(x)sin(x)dx$

Then I'm kind of stuck... my next guess would be to take this identity $sin(2x) = 2sin(x)cos(x)$

2. Originally Posted by larson
$\int sec(x)sin^3(x)dx$

$\int \frac {tan(x)}{sin(x)} sin^3(x)dx$

So then the sin(x)'s cancel out leaving...
[tex] \int tan(x)sin^2(x)dx

then

$\int \frac {sin(x)}{cos(x)} sin^2(x)dx$

this equals

$\int \frac {sin^3(x)}{cos(x)}dx$

Then split up the $sin^3(x)$ and then set the $sin^2(x)$ equal to $1 - cos^2(x)$

$\int \frac {1 - cos^2(x)}{cos(x)} sin(x)dx$

Then the cos(x)'s can cancel...

$\int (1 - cos(x)) sin(x)dx$

this then equals

$\int sin(x) - cos(x)sin(x)dx$

Then I'm kind of stuck... my next guess would be to take this identity $sin(2x) = 2sin(x)cos(x)$
for $\int \cos x \sin x~dx$, why not just do a simple substitution? $u = \sin x$ or $u = \cos x$ will suffice

by the way, you could have went from $\int \sec x \sin^3 x~dx$ directly to $\int \frac {\sin^3 x}{\cos x}~dx$

but wait...you "canceled" the cosines incorrectly, you can't cancel through a sum like that.

from $\int \frac {\sin^3 x}{\cos x}~dx$ make the substitution $u = \cos x$

3. Hello, larson

Another approach . . .

$\int \sec x \sin^3\!x \,dx$

We have: . $\frac{\sin^3\!x}{\cos x} \;=\;\frac{\sin^2x}{\cos x}(\sin x) \;=\;\frac{1-\cos^2\!x}{\cos x}(\sin x)$

Then we have: . $\int \left(\frac{1}{\cos x} - \cos x\right)(\sin x\,dx)$

Let $u \,=\,\cos x\quad\Rightarrow\quad du \,=\,-\sin x\,dx\quad\Rightarrow\quad \sin x\,dx \,=\,-du$

Substitute: . $\int\left(\frac{1}{u} - u\right)(-du) \;=\;\int\left(u - \frac{1}{u}\right)\,du \;=\;\frac{1}{2}u^2 - \ln|u| + C$

Back-substitute: . $\frac{1}{2}\cos^2\!x - \ln|\cos x| + C$

4. Originally Posted by Soroban
Hello, larson

Another approach . . .

We have: . $\frac{\sin^3\!x}{\cos x} \;=\;\frac{\sin^2x}{\cos x}(\sin x) \;=\;\frac{1-\cos^2\!x}{\cos x}(\sin x)$

Then we have: . $\int \left(\frac{1}{\cos x} - \cos x\right)(\sin x\,dx)$

Let $u \,=\,\cos x\quad\Rightarrow\quad du \,=\,-\sin x\,dx\quad\Rightarrow\quad \sin x\,dx \,=\,-du$

Substitute: . $\int\left(\frac{1}{u} - u\right)(-du) \;=\;\int\left(u - \frac{1}{u}\right)\,du \;=\;\frac{1}{2}u^2 - \ln|u| + C$

Back-substitute: . $\frac{1}{2}\cos^2\!x - \ln|\cos x| + C$

Thank you .