# Thread: Integration with Trig. Sub.

1. ## Integration with Trig. Sub.

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.

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

because of the pythagorean identity ...

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

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

4. 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.
$\displaystyle -\int\frac{5x}{(x^2+5)^{3/2}}\,dx$

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

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

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

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

Putting it all together, we have:

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

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

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

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