Integral [(cosx)^3-(sinx)^2]dx/(cosx)2

Can i cancel the cosx leaving cosx in numerator before I integrate?

If anyone could help with integrating this problem -

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- Jul 30th 2008, 05:25 PMweezie23Integration
Integral [(cosx)^3-(sinx)^2]dx/(cosx)2

Can i cancel the cosx leaving cosx in numerator before I integrate?

If anyone could help with integrating this problem - - Jul 30th 2008, 05:29 PMChris L T521
You have the right idea.

Split up the integrand:

$\displaystyle \frac{\cos^3x-\sin^2x}{\cos^2x}=\frac{\cos^3x}{\cos^2x}-\frac{\sin^2x}{\cos^2x}=\cos x-\tan^2x$

Now can you integrate $\displaystyle \int \cos x-\tan^2x\,dx$?

*hint*: $\displaystyle \tan^2x=\sec^2x-1$

I hope this clarifies things! (Sun)

--Chris - Jul 30th 2008, 05:59 PMweezie23
- Jul 30th 2008, 06:02 PMChris L T521
- Jul 30th 2008, 06:03 PMweezie23or
or i should say sinx-tanx + c

- Jul 30th 2008, 06:05 PMweezie23or
sinx-tanx -x +c

- Jul 30th 2008, 06:06 PMChris L T521
- Jul 30th 2008, 06:08 PMweezie23last time
sinx-tanx+x+c

- Jul 30th 2008, 06:08 PMChris L T521
- Jul 30th 2008, 06:09 PMChris L T521