Mega hard integral part 3 <plz help>

[caseclosed]

I hope I am not bothering you guys with too many questions and post

I am clueless, tried changing it to csc^2(x)csc(x) but doesn't help neither does csc(x)(1+cot^2(x))

[Tom Mattson]

Try integrating by parts. Let [imath]u=\csc(x)[/imath] and let [imath]du=\csc^2(x)dx[/imath].

[caseclosed]

Try integrating by parts. Let [imath]u=\csc(x)[/imath] and let [imath]du=\csc^2(x)dx[/imath'].

I tried that and I did

first time- U=csc(x) du=-csc(x)cot(x)dx dv=csc^2(x)dx v=-cot(x)

second time - U=cot(x) du=-csc^2(x)dx dv=csc(x)cot(x)dx v=-csc(x)

 

after doing all that it loops back to the problem so I get -csc(x)cot(x)+cot(x)csc(x)+I=I where I=integral of csc^3(x)dx

so I subtract I, I get 0....

 

are the Us an Dv's suppose to stay consistent?

[Tom Mattson]

That is the correct method, but you should not have gotten 0. You must have made a small sign error.

[caseclosed]

here is my work, I don't see anywhere that I made the wrong sign.

-csc(x)cot(x)-integral(-csc(x)(-cot^2(x)dx)

-csc(x)cot(x)-[-cot(x)csc(x)-integral(-csc(x)(-csc^2(x))dx]

-csc(x)cot(x)-cot(x)csc(x)+integral(csc^3(x)dx)

so let integral(csc^3(x)dx)=I

I get -csc(x)cot(x)-cot(x)csc(x)+I=I

....

[Tom Mattson]

here is my work' date=' I don't see anywhere that I made the wrong sign.

-csc(x)cot(x)-integral(-csc(x)(-cot^2(x)dx)

[/quote']

 

OK.

 

-csc(x)cot(x)-[-cot(x)csc(x)-integral(-csc(x)(-csc^2(x))dx]

 

Not sure how this came about. In your previous step you should have inserted [imath]\cot^2(x)=\csc^2(x)-1[/imath] to obtain the following.

 

[math]\int\csc^3(x)dx=-\csc(x)\cot(x)-\int\csc^3(x)dx+\int\csc(x)dx[/math]

 

It all works out straightforwardly from there.

[caseclosed]

awww, I did parts twice that is why did not work for me.