# Integration

• Jul 30th 2008, 05:25 PM
weezie23
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 -
• Jul 30th 2008, 05:29 PM
Chris L T521
Quote:

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:

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

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

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

I hope this clarifies things! (Sun)

--Chris
• Jul 30th 2008, 05:59 PM
weezie23
integrate = sinx-tanx-1

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

Quote:

Originally Posted by Chris L T521
You have the right idea.

Split up the integrand:

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

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

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

I hope this clarifies things! (Sun)

--Chris

• Jul 30th 2008, 06:02 PM
Chris L T521
Quote:

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
• Jul 30th 2008, 06:03 PM
weezie23
or
or i should say sinx-tanx + c
• Jul 30th 2008, 06:05 PM
weezie23
or
sinx-tanx -x +c
• Jul 30th 2008, 06:06 PM
Chris L T521
Quote:

Originally Posted by weezie23
or i should say sinx-tanx + c

Not quite...

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

Fill in the blank...

--Chris
• Jul 30th 2008, 06:08 PM
weezie23
last time
sinx-tanx+x+c
• Jul 30th 2008, 06:08 PM
Chris L T521
Quote:

Originally Posted by weezie23
sinx-tanx -x +c

Make it plus, and then you have it

--Chris
• Jul 30th 2008, 06:09 PM
Chris L T521
Quote:

Originally Posted by weezie23
sinx-tanx+x+c

Good. You have it now.

--Chris