[SOLVED] Trigonometric integration by substitution

• Jul 2nd 2008, 08:36 PM
redman223
[SOLVED] Trigonometric integration by substitution
$\int \frac {1-tanx^2}{secx^2} ~dx$

I figure you have to break it up first:

$\int cosx^2 ~dx - \int \frac {tanx^2}{secx^2} ~dx$

But from here, I am somewhat stuck. Would $\int \frac {tanx^2}{secx^2} ~dx$ turn into $\int sinx^2 ~dx$?

If so, then I guess the next part of the question is, how do you take the anti derivative of $cosx^2$?
• Jul 2nd 2008, 08:48 PM
mr fantastic
Quote:

Originally Posted by redman223
$\int \frac {1-tanx^2}{secx^2} ~dx$

I figure you have to break it up first:

$\int cosx^2 ~dx - \int \frac {tanx^2}{secx^2} ~dx$

But from here, I am somewhat stuck. Would $\int \frac {tanx^2}{secx^2} ~dx$ turn into $\int sinx^2 ~dx$?

If so, then I guess the next part of the question is, how do you take the anti derivative of $cosx^2$?

Yes.

$\cos (2x) = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 (x) - 1$

Therefore $\sin^2 x = \frac{1 - \cos (2x)}{2}$ and $\cos^2 x = \frac{\cos (2x) + 1}{2}\,$ ......

But of course, you could always recognise that $\frac{1- \tan^2 x}{\text{sec} ^2 x} = \cos^2 x - \sin^2 x = \cos (2x)$ and save yourself a lot of trouble ......
• Jul 2nd 2008, 08:57 PM
Msquared
Quote:

Originally Posted by redman223
$\int \frac {1-tanx^2}{secx^2} ~dx$

I figure you have to break it up first:

$\int cosx^2 ~dx - \int \frac {tanx^2}{secx^2} ~dx$

But from here, I am somewhat stuck. Would $\int \frac {tanx^2}{secx^2} ~dx$ turn into $\int sinx^2 ~dx$?

If so, then I guess the next part of the question is, how do you take the anti derivative of $cosx^2$?

$\int \frac {1-tanx^2}{secx^2} ~dx$

Are you sure its that integral or this?

$
\int \frac {1-(tanx)^2}{(secx)^2} ~dx
$

Because if its the first, then its not going to be simple.(I was kindof thrown off by the X^2 part)
• Jul 2nd 2008, 09:08 PM
Mathstud28
Quote:

Originally Posted by Msquared
$\int \frac {1-tanx^2}{secx^2} ~dx$

Are you sure its that integral or this?

$
\int \frac {1-(tanx)^2}{(secx)^2} ~dx
$

Because if its the first, then its not going to be simple.(I was kindof thrown off by the X^2 part)

$\frac{1-\tan^2(x)}{\sec^2(x)}=\frac{2-(\tan^2(x)+1)}{\tan^2(x)+1}=\frac{2}{\sec^2(x)}-1=2\cos^2(x)-1=\cos(2x)$

EDIT: Oops! Looks as though I am steping on Mr. F's toes again...desole!
• Jul 2nd 2008, 09:08 PM
redman223
Its the second one, not the first one. It is not being evaluated at x^2.

However, I am confused as to why:
$\cos^2 x - \sin^2 x = \cos (2x)$

I keep getting $2cos^2x -1$

$cos^2x - (1-cos^2x)$
• Jul 2nd 2008, 09:11 PM
Mathstud28
Quote:

Originally Posted by redman223
Its the second one, not the first one. It is not being evaluated at x^2.

However, I am confused as to why:
$\cos^2 x - \sin^2 x = \cos (2x)$

I keep getting $2cos^2x -1$

$cos^2x - (1-cos^2x)$

$2\cos^2(x)-1=\cos(2x)$

$2\cdot\frac{e^{2ix}+2+e^{-2ix}}{4}-1=\frac{e^{2ix}+2+e^{-2ix}}{2}-1$ $=\frac{e^{2ix}+2+e^{-2ix}-2}{2}=\frac{e^{2ix}+e^{-2ix}}{2}=\cos(2x)\quad\blacksquare$
• Jul 2nd 2008, 09:28 PM
redman223
Thanks for the help, my next question is:

$\int cos(2x) ~dx = \frac {sin(2x)}{2}$ ?
• Jul 2nd 2008, 09:32 PM
Mathstud28
Quote:

Originally Posted by redman223
Thanks for the help, my next question is:

$\int cos(2x) = \frac {sin(2x)}{2}$ ?

$\cdots+\bold{\color{red}C}$
• Jul 2nd 2008, 09:55 PM
mr fantastic
Quote:

Originally Posted by redman223
Lol, I always forget that.

On to the next question. Using $x=3sin\theta$

$\int x^3 \sqrt {9-x^2} ~dx$

I think I solved the part under the radical correctly ( $3cos\theta$)

If I plug in $x=3sin\theta$ and $dx=3cos\theta$ and use $3cos\theta$ for the radical I get:

$\int 243sin^3\theta cos^2\theta ~d\theta$

Thats where I get lost.

New and unrelated questions require new threads. Delete this question (copy it first!) and re-post it as a new thread.

Edit: Belay that request - I see Mathstud's doing more stepping. But note it for next time.
• Jul 2nd 2008, 09:56 PM
Mathstud28
Quote:

Originally Posted by redman223
Lol, I always forget that.

On to the next question. Using $x=3sin\theta$

$\int x^3 \sqrt {9-x^2} ~dx$

I think I solved the part under the radical correctly ( $3cos\theta$)

If I plug in $x=3sin\theta$ and $dx=3cos\theta$ and use $3cos\theta$ for the radical I get:

$\int 243sin^3\theta cos^2\theta ~d\theta$

Thats where I get lost.

$\sin^3(\theta)\cos^2(\theta)=\sin(\theta)(1-\cos^(\theta))(\cos^2(\theta))$

Now expand
• Jul 2nd 2008, 10:06 PM
mr fantastic
Getting off-topic but a legitimate response I think:

Quote:

Originally Posted by Mathstud28;163721[snip
EDIT: Oops! Looks as though I am steping on Mr. F's toes again...desole!

Like the song says:

Je suis désolé, mais je n'ai pas le choix
Je suis désolé, mais la vie me demande ça

By the way ...... My toes are on my feet. My feet are usually in my mouth. So watch the face, mathstud.

How's that for setting up the (obscure) double entendre for desole .....?
• Jul 2nd 2008, 10:12 PM
Mathstud28
Quote:

Originally Posted by mr fantastic
New and unrelated questions require new threads. Delete this question (copy it first!) and re-post it as a new thread.

Edit: Belay that request - I see Mathstud's doing more stepping. But note it for next time.

What does stepping mean in this context? (Thinking)
• Jul 2nd 2008, 10:13 PM
Mathstud28
Quote:

Originally Posted by mr fantastic

How's that for setting up the (obscure) double entendre for desole .....?

Acceptable (Nod)