Integration with Trig. Sub.

**jlmills5** · October 11th 2009, 10:24 AM

The question-

int [(-5x)/(x^2 + 5)^(3/2)] dx

I have-

x=sqrt(5)tan(theta)
dx=sqrt(5)sec^2(theta)dtheta

my solutions manual says that (x^2 + 5) = 5sec^2(theta) <---where does this come from. Is there some trig identity I'm not seeing?

I've attached an image which may show the problem better.

**skeeter** · October 11th 2009, 10:43 AM

Originally Posted by jlmills5

The question-

int [(-5x)/(x^2 + 5)^(3/2)] dx

I have-

x=sqrt(5)tan(theta)
dx=sqrt(5)sec^2(theta)dtheta

my solutions manual says that (x^2 + 5) = 5sec^2(theta) <---where does this come from. Is there some trig identity I'm not seeing?

I've attached an image which may show the problem better.

$5\tan^2{t} + 5 = 5(\tan^2{t} + 1) = 5\sec^2{t}$

because of the pythagorean identity ...

$\tan^2{t} + 1 = \sec^2{t}$

**jlmills5** · October 11th 2009, 10:45 AM

Ah. I knew it would be something simple I wasn't seeing. Thank you!!!

**redsoxfan325** · October 11th 2009, 10:51 AM

Originally Posted by jlmills5

The question-

int [(-5x)/(x^2 + 5)^(3/2)] dx

I have-

x=sqrt(5)tan(theta)
dx=sqrt(5)sec^2(theta)dtheta

my solutions manual says that (x^2 + 5) = 5sec^2(theta) <---where does this come from. Is there some trig identity I'm not seeing?

I've attached an image which may show the problem better.

$-\int\frac{5x}{(x^2+5)^{3/2}}\,dx$

$x=\sqrt{5}\tan(u) \implies dx=\sqrt{5}\sec^2(u)\,du$

Using the identity $1+\tan^2(\alpha)=\sec^2(\alpha)$ ,

We have that $(\sqrt{5}\tan(u))^2+5=5\tan^2(u)+5=5\sec^2(u)$ .

$(5\sec^2(u))^{3/2}=5\sqrt{5}\sec^3(u)$

Putting it all together, we have:

$-\int\frac{5\sqrt{5}\tan(u)}{5\sqrt{5}\sec^3(u)}\cd ot\sqrt{5}\sec^2(u)\,du = -\sqrt{5}\int\frac{\tan(u)}{\sec(u)}\,du=-\sqrt{5}\int\sin(u)\,du=\sqrt{5}\cos(u)$

Plug back in for $u$ and you're done.

**redsoxfan325** · October 11th 2009, 12:06 PM

Originally Posted by jlmills5

The question-

int [(-5x)/(x^2 + 5)^(3/2)] dx

I have-

x=sqrt(5)tan(theta)
dx=sqrt(5)sec^2(theta)dtheta

my solutions manual says that (x^2 + 5) = 5sec^2(theta) <---where does this come from. Is there some trig identity I'm not seeing?

I've attached an image which may show the problem better.

However, it would be much simpler to do a regular u-sub (without using trig):

Let $u=x^2+5 \implies du=2x\,dx$

$-\frac{5}{2}\int\frac{du}{u^{3/2}}=-\frac{5}{2}\cdot-2\cdot\frac{1}{u^{1/2}}=\frac{5}{\sqrt{u}}=\boxed{\frac{5}{\sqrt{x^2+5 }}}$

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