Integration with Trig. Substitution

**flybynight** · September 19th 2010, 02:04 PM

Hey guys:
Had one problem with my math homework:

$\int \frac{y^2}{25+y^2} dy$

So I tried $y=5\tan x$

So then: $dy=5\sec^2xdx$ ,

So the integration appears as: $\int \frac{25\tan^2x}{25(1+\tan^2x)}(5\sec^2x)dx$

Which simplifies to: $\int 5\tan^2x dx$ , which is where I get stuck.

Any advice on moving forward, or a different, easier method?

Peter

**mr fantastic** · September 19th 2010, 02:23 PM

Originally Posted by flybynight

Hey guys:
Had one problem with my math homework:

$\int \frac{y^2}{25+y^2} dy$

So I tried $y=5\tan x$

So then: $dy=5\sec^2xdx$ ,

So the integration appears as: $\int \frac{25\tan^2x}{25(1+\tan^2x)}(5\sec^2x)dx$

Which simplifies to: $\int 5\tan^2x dx$ , which is where I get stuck.

Any advice on moving forward, or a different, easier method?

Peter

Make life easy for yourself by noting that $\displaystyle \frac{y^2}{y^2 + 25} = \frac{(y^2 + 25) - 25}{y^2 + 25} = 1 - \frac{1}{y^2 + 25}$ .

On the other hand, if you're determined to know how to find $\displaystyle \int \tan^2 (x) \, dx$ then you should note that $\tan^2 (x) = \sec^2 (x) - 1$ (at which point it helps to know what the derivative of $\tan (x)$ is).

**flybynight** · September 19th 2010, 03:56 PM

Here's what I did:
$\int 1-\frac{25}{y^2+25} dy$

$y=5 \tan(x)$

$dy=5 \sec^2(x) dx$

$\int 1-\frac{25}{25(\tan^2(x))} 5 \sec^2(x) dx$

$\int 1-\frac{5\sec^2(x)}{\sec^2(x)} dx$

$\int 1-5 dx$

$-4y=-4\arctan(\frac{y}{5})$

However, the answer in the textbook (confirmed by Mathematica) is $y-5\arctan(\frac{y}{5})$

Any help?

Peter

**Defunkt** · September 19th 2010, 04:45 PM

Your algebra is a bit messy.

First, use linearity to pull out the 1 term from the integral and get

$\displaystyle I = \int \left( 1 - \frac{25}{y^2 + 25} \right) dy = \int 1dy - 25 \int \frac{dy}{y^2 + 25} = \int 1dy - \frac{25}{25} \int \left( \frac{1}{tan^2x + 1} \right) \cdot 5sec^2xdx$

Remembering that $tan^2x + 1 = sec^2x$ and thus $\frac{1}{tan^2x + 1} = cos^2x$ you get

$\displaystyle I = y - \int 5dx = y - 5x = y - 5tan^{-1} \left(\frac{y}{5} \right)$

Your mistake was that from the first integral to the second (after performing the substitution), you multiplied only the $\frac{25}{y^2+25}$ factor by $5sec^2xdx$ instead of the whole integrand. You also forgot a +25 in the denominator there.

**flybynight** · September 19th 2010, 05:04 PM

Oh I see. I can't believe that I over looked that. Thank you very much!

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