Question

Can someone check this integral for me?

Answer 1

wolframalpha.com

Answer 2

oh stop it @jk_16

Answer 3

stop it what..wolfram works

Answer 4

The asker is probably looking for a way to learn it instead of a flat out answer.

Answer 5

This^

Answer 6

^

Answer 7

lol kids..wolfram does just fine with the work..if u dont understand a part the solution, this is what we are here for

Answer 8

lol @ kids who use wolfram

Answer 9

\[\int\limits_{}^{}\csc^3x\]using Integration by parts

u=cscx dv=csc^2x
du=-cscxcotx v=-cotx

\[-cscxcotx-\int\limits_{}^{}cscxcot^2xdx\]

and then substitution with

w=cotx dw=cscxcotxdx

\[-cscxcotx-\int\limits_{}^{}-w dw = -cscxcotx+1/2 w^2 + c\]and a final answer of \[\int\limits_{}^{}\csc^3x = -cscxcotx+1/2\cot^2x+c\]

Answer 10

Should be - (1/2)cot x csc x + (1/2) ln | csc x - cot x | + C

Answer 11

If w = cotx then dw = -csc^2

Answer 12

I think...

Answer 13

@cgreenwade2000 It's csc^3(x)

Answer 14

@Dido525 Right the remaining integral would be csc^3

Answer 15

do u know what the reduction formula for integral csc^m(x) dx is

Answer 16

Dido your answer above is the one in the back of the book, where did I go wrong from getting that answer?

Answer 17

Hint: Use a trig identity for csc^2(x) when you intergrate by parts.

Answer 18

Rewrite as ∫ csc²x csc x dx

Answer 19

Now try again?

Answer 20

csc²x dx = dv

- cot x = v

csc x = u

- cot x csc x dx = du

Answer 21

Can we stop this whole thing and have him rewrite the problem by fixing his second substitution?

Answer 22

Okay.

Answer 23

So that second substitution is bogus seeing as I somehow decided the derivative of cot was csccot. Anyways, without that substitution I am still stuck on where to go from\[-cscxcotx-\int\limits_{}^{}cscxcot^2(x)\]Any suggestions?