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Hard integral

I have this integral and the answer is not the same as on the calculator but im not sure if its right or wrong.
∫1/cos^2(x)(tan^3(x)-tan(x)) . I simplified it down to ∫sin(2x)/cos^2((2x)-1 . I then substituted cos(2x) and the integral came out eventually as (ln(-cot^2(x)))/4
I'm not sure that's right or how to really check it?

Reply 1

ghostwalker

Original post by yusyus

Given that your answer takes the ln of a negative number, it's not going to be correct. Check for mistakes, and/or post your full working.

Reply 2

yusyus

Original post by ghostwalker

Given that your answer takes the ln of a negative number, it's not going to be correct. Check for mistakes, and/or post your full working.

Hi, this is my working I did make a mistake originally but I'm not sure how to check my answer now

hard integral.jpg66.8KB

Reply 3

yusyus

Original post by TeeEm

Try the substitution u=tanx from the start

ahh, that does seem like a much easier method now, I just want to see if my original method is right though. Thanks!

Reply 4

ghostwalker

Original post by yusyus

I just want to see if my original method is right though. Thanks!

Your log manipulation at the end has a rather basic error.

You seem to have done essentially, ln(a-b) = ln(a) - ln(b).

Reply 5

yusyus

Original post by ghostwalker

Your log manipulation at the end has a rather basic error.

You seem to have done essentially, ln(a-b) = ln(a) - ln(b).

I guess I can blame that on my lack of practice - thanks. Is it right before that point though?

Reply 6

ghostwalker

Original post by yusyus

I guess I can blame that on my lack of practice - thanks. Is it right before that point though?

Didn't spot any errors.

As a check, you can differentiate to see if you get back to your original function. Rather tedious in this case; I'd recommend using Wolfram Alpha (just google it). It takes a little while to pick up the syntax/vocab, but worth the effort. Easiest to check out their examples when starting out.