Integrals involving trig functions

[Chandasouk]

Homework Statement

I need help evaluating the integral Cotx³/10

I factored out the 1/10 from the integral and am just left with (1/10)*Cotx³

from here i do not really know what to do. I rewrote it in terms of sine and cosine to get

(1/10)*(Cosx³/Sinx³)dx

I multiply the integral by (1/Sinx³) to get rid of the denominator and am left with

(1/10)*(Cosx³dx

I factor out a (Cosx²) and am left with

(1/10)*(Cosx²)(Cosx)dx

Rewriting using trig indentities, I get

(1/10)*(1-Sinx²)(Cosx)dx

rewriting i get

(1/10)*(cosx-sinx²cosx)dx

I solve this integral and get

(1/10)sinx - (1/10)*((sinx)³/3) + C

but that is incorrect. The answer is supposed to be

(-1/20)cotx²-(1/10)ln(sinx)+C

What went wrong?

 

[vela]

Well, for one thing, you can't just get rid of parts of the integrand.

I multiply the integral by (1/Sinx³) to get rid of the denominator and am left with

(1/10)*(Cosx³dx

Just to be clear, you are trying to compute

[tex]\int \cot^3 x\,dx[/tex]

and not

[tex]\int \cot x^3\, dx[/tex]

right?

Hint: Try the substitution u=sin x.

 

[Chandasouk]

Yes, Cot³x

 

[HallsofIvy]

And is it
[tex]\frac{cot^3 (x)}{10}[/tex]
or
[tex]cot^3\left(\frac{x}{10}\right)[/tex]
?

Try writing [itex]cot^3(x)[/itex] as

[tex]\frac{sin^3(x)}{cos^3(x)}= [/tex][tex]\frac{sin^2(x)}{cos^3(x)}cos(x)=[/tex][tex] \frac{1- cos^2(x)}{cos^3(x)} sin(x)[/tex]

and use the substitution u= cos(x).

 

[Chandasouk]

And is it
[tex]\frac{cot^3 (x)}{10}[/tex]
or
[tex]cot^3\left(\frac{x}{10}\right)[/tex]
?

Try writing [itex]cot^3(x)[/itex] as

[tex]\frac{sin^3(x)}{cos^3(x)}= [/tex][tex]\frac{sin^2(x)}{cos^3(x)}cos(x)=[/tex][tex] \frac{1- cos^2(x)}{cos^3(x)} sin(x)[/tex]

and use the substitution u= cos(x).

It is [tex]\frac{cot^3 (x)}{10}[/tex]

I actually have no idea how to use most of the math tags, so writing it out is hard.

Cot³x is [tex]\frac{Cos^3(x)}{Sin^3(x)}[/tex]

I'll try distributing out a cosine factor and converting the rest to sign and see how it goes