1. ## Integration

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 -

2. Originally Posted by weezie23
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 -
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!

--Chris

3. integrate = sinx-tanx-1

b/c sec^2x -1 = tan^2x and Int.sec^2xdx= tanx+c

Originally Posted by Chris 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!

--Chris

4. Originally Posted by weezie23
integrate = sinx-tanx+1

b/c sec^2x -1 = tan^2x and Int.sec^2xdx= tanx+c
VERY close...look at what I highlighted in red. First, it should be plus, not minus. Secondly, do you know what you forgot to do to the last term? Also, you forgot the constant of integration...

--Chris

5. ## or

or i should say sinx-tanx + c

6. ## or

sinx-tanx -x +c

7. Originally Posted by weezie23
or i should say sinx-tanx + c
Not quite...

$\displaystyle \int \cos x-\sec^2x+1\,dx=\sin x-\tan x+\text{ }+C$

Fill in the blank...

--Chris

8. ## last time

sinx-tanx+x+c

9. Originally Posted by weezie23
sinx-tanx -x +c
Make it plus, and then you have it

--Chris

10. Originally Posted by weezie23
sinx-tanx+x+c
Good. You have it now.

--Chris