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

kpx001

Junior Member
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Mar 6, 2006
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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
 
Re: integrals

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

Hello, kpx001!

We have:   1sin(x1)cos(x1)sin(x1)dx  =  csc(x1)cot(x1)dxGot it?\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?}


(secx)32tanxdx\displaystyle \int (\sec x)^{\frac{3}{2}}\tan x\,dx
We have:   (secx)12(secxtanxdx)\displaystyle \text{We have: }\; \int (\sec x)^{\frac{1}{2}}(\sec x\tan x\,dx)

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

Substitute:   u12duetc.\displaystyle \text{Substitute: }\;\int u^{\frac{1}{2}}du \quad \cdots \quad \text{etc.}


sin3 ⁣xcos3 ⁣xdx\displaystyle \int \sin^3\!x\cos^3\!x\,dx

We have:   sin3 ⁣xcos2 ⁣x(cosxdx)  =  sin3 ⁣x(1sin2 ⁣x)(cosxdx)  =  (sin3 ⁣xsin5 ⁣x)(cosxdx)\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)

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

Substitute:   (u3u5)duetc.\displaystyle \text{Substitute: }\;\int(u^3-u^5)\,du\quad\cdots\quad \text{etc.}

 
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