[SOLVED] Trigonometric integration by substitution

**redman223** · July 2nd 2008, 07:36 PM

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

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 $cosx^2$ ?

**mr fantastic** · July 2nd 2008, 07:48 PM

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 ......

**Msquared** · July 2nd 2008, 07:57 PM

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?

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

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

**Mathstud28** · July 2nd 2008, 08:08 PM

Originally Posted by Msquared

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

Are you sure its that integral or this?

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

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!

**redman223** · July 2nd 2008, 08:08 PM

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 $2cos^2x -1$

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

**Mathstud28** · July 2nd 2008, 08:11 PM

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 $2cos^2x -1$

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

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

$2\cdot\frac{e^{2ix}+2+e^{-2ix}}{4}-1=\frac{e^{2ix}+2+e^{-2ix}}{2}-1$ $=\frac{e^{2ix}+2+e^{-2ix}-2}{2}=\frac{e^{2ix}+e^{-2ix}}{2}=\cos(2x)\quad\blacksquare$

**redman223** · July 2nd 2008, 08:28 PM

Thanks for the help, my next question is:

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

**Mathstud28** · July 2nd 2008, 08:32 PM

Originally Posted by redman223

Thanks for the help, my next question is:

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

$\cdots+\bold{\color{red}C}$

**mr fantastic** · July 2nd 2008, 08:55 PM

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.

**Mathstud28** · July 2nd 2008, 08:56 PM

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

**mr fantastic** · July 2nd 2008, 09:06 PM

Getting off-topic but a legitimate response I think:

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 .....?

**Mathstud28** · July 2nd 2008, 09:12 PM

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?

**Mathstud28** · July 2nd 2008, 09:13 PM

Originally Posted by mr fantastic

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

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