integrals: cos(x-1)dx/(sin(x-1))^2, etc.

[kpx001]

im having a bit of trouble solving these integrals

1. integral of cos(x-1)dx/(sin(x-1))^2

i was thinking a u sub for the denominator but it has a ^2

2. integral of (secx)^3/2 * tanxdx

i tried rewriting sec as cos(x)^3/2 * sinx/cosxdx

3. integral of (sinx)^3 * (cosx)^3dx
(sinx)^3(cosx)^3(cosx)^3
(sinx)^3(1-(sinx)^2)cosx = (sinx)^3 - (sinx)^5 * (cosx)
(sinx)^3* cosx - (sinx)^5 * cosx
(u^3-u^5)du
and im stuck here

 

[royhaas]

Re: integrals

The substitution works fine for the first integral. For the second one, use \(\displaystyle y=sec(x)\).

 

[soroban]

Re: integrals

Hello, kpx001!

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

\(\displaystyle \text{We have: }\;\int\frac{1}{\sin(x-1)}\cdot\frac{\cos(x-1)}{\sin(x-1)}\,dx \;=\;\int\csc(x-1)\cot(x-1)\,dx \quad \cdots\quad \text{Got it?}\)

\(\displaystyle \int (\sec x)^{\frac{3}{2}}\tan x\,dx\)

\(\displaystyle \text{We have: }\; \int (\sec x)^{\frac{1}{2}}(\sec x\tan x\,dx)\)

\(\displaystyle \text{Let: }\,u = \sec x\quad\Rightarrow\quad du = \sec x\tan x\,dx\)

\(\displaystyle \text{Substitute: }\;\int u^{\frac{1}{2}}du \quad \cdots \quad \text{etc.}\)

\(\displaystyle \int \sin^3\!x\cos^3\!x\,dx\)

\(\displaystyle \text{We have: }\;\int\sin^3\!x \cos^2\!x(\cos x\,dx) \;=\;\int\sin^3\!x(1-\sin^2\!x)(\cos x\,dx) \;=\;\int(\sin^3\!x - \sin^5\!x)(\cos x\,dx)\)

\(\displaystyle \text{Let: }\:u \:= \:\sin x\quad\Rightarrow\quad du \:=\:\cos x\,dx\)

\(\displaystyle \text{Substitute: }\;\int(u^3-u^5)\,du\quad\cdots\quad \text{etc.}\)