# Thread: trig substitutions for integrals

1. ## trig substitutions for integrals

i have a test coming up and i wanted to double-check my answers:

1. integral (b = 1, a = 0) x(sqrt (x^2 + 4)) dx
i got 1/8 - 1/4sqrt(5) + c

2. integral (du)/(u(sqrt(5-u^2)))
1/5 ln u/sqrt(5) ln (sqrt(5) + u)/sqrt(5-^2)) + c

3 integral (dt)/(sqrt(t^2 -6t +13))
1/3 arctan (x-3)/2 + c

if someone could tell me if i got these right...it would help me so muchh..thanks

2. Hello !
Originally Posted by skabani
i have a test coming up and i wanted to double-check my answers:

1. integral (b = 1, a = 0) x(sqrt (x^2 + 4)) dx
i got 1/8 - 1/4sqrt(5) + c
Nope ! There is a problem here !

Substitute $\displaystyle y=x^2+4$

after some working, we have :

$\displaystyle =\frac 12 \int_4^5 y^{1/2} ~dx$ (remember, $\displaystyle \sqrt{a}=a^{1/2}$

But the primitive of $\displaystyle y^n$ is $\displaystyle \frac{y^{n+1}}{n+1}$. Here, n=1/2 ---> n+1=3/2 ---> 1/(n+1)=2/3

2. integral (du)/(u(sqrt(5-u^2)))
1/5 ln u/sqrt(5) ln (sqrt(5) + u)/sqrt(5-^2)) + c

3 integral (dt)/(sqrt(t^2 -6t +13))
1/3 arctan (x-3)/2 + c
It's 1/2 ...

or any mathematical software or calculator !

3. okay,

so is number one: 111/5

two, i re-did it and got: (1/sqrt(5)) ln|csc x - cot x|+ c
i know we have to substitute the original variables back into the answer to get it in terms of u, but im not sure how to do that..

three: im not sure how to get 1/2, do you mind showing me

i tried the website, but all it does is factors out the variable and gives me the original integral...thanks anyways though

4. For the second problem, you should not be getting any natural logs in the answer.

The integral is:
$\displaystyle \int \frac{u}{\sqrt{5 - u^2}} ~ du$

By multiplying by -1 inside and -1 outside you get

$\displaystyle -\int \frac{-u}{\sqrt{5 - u^2}} ~ du$

and the integral evaluates to
$\displaystyle -\sqrt{5 - u^2} + C$.

5. Originally Posted by icemanfan
For the second problem, you should not be getting any natural logs in the answer.

The integral is:
$\displaystyle \int \frac{u}{\sqrt{5 - u^2}} ~ du$

By multiplying by -1 inside and -1 outside you get

$\displaystyle -\int \frac{-u}{\sqrt{5 - u^2}} ~ du$

and the integral evaluates to
$\displaystyle -\sqrt{5 - u^2} + C$.
i thought we were supposed to use trig substitution- set u=sqrt(5) sin x
and solve from there?

6. Originally Posted by skabani
i thought we were supposed to use trig substitution- set u=sqrt(5) sin x
and solve from there?
You can use trig substitution sometimes when you have $\displaystyle \frac{1}{\sqrt{a^2 - x^2}}$ or similar situations, but in this case, the integral is in the $\displaystyle y^n ~ dy$ form, where $\displaystyle y = 5 - u^2, n = -\frac{1}{2}, dy = -2u ~ du$. It is easier to use the $\displaystyle y^n ~ dy$ form if you can.

7. Originally Posted by icemanfan
You can use trig substitution sometimes when you have $\displaystyle \frac{1}{\sqrt{a^2 - x^2}}$ or similar situations, but in this case, the integral is in the $\displaystyle y^n ~ dy$ form, where $\displaystyle y = 5 - u^2, n = -\frac{1}{2}, dy = -2u ~ du$. It is easier to use the $\displaystyle y^n ~ dy$ form if you can.
oh, okay, and do you mind checking number 1 for me..i got 111/5

thanks so much!!

8. I thought the second integral was:
$\displaystyle \int \frac{du}{u\sqrt{5-u^2}}$

From what skabani wrote in his first post, it looks that way. Can you clear this up Skabani?

9. I'm confused is the second integral

1st form $\displaystyle \int \frac{du}{u\sqrt{5-u^2}}$

2nd form $\displaystyle \int \frac{u~du}{\sqrt{5-u^2}}$

10. Originally Posted by Chop Suey
I thought the second integral was:
$\displaystyle \int \frac{du}{u\sqrt{5-u^2}}$

From what skabani wrote in his first post, it looks that way. Can you clear this up Skabani?
Oops, I think you're right. I read the integral wrong. Looks like trig substitution is the way to go.

11. Originally Posted by Chop Suey
I thought the second integral was:
$\displaystyle \int \frac{du}{u\sqrt{5-u^2}}$

From what skabani wrote in his first post, it looks that way. Can you clear this up Skabani?

actually, you are right, it is what chop suey said...

12. Then substitute $\displaystyle t = \frac{1}{\sqrt{5-u^2}}$ and see where that gets you.

13. O ok well then you can approach it like this

divide top and bottom by $\displaystyle \sqrt{5}$

which leaves

$\displaystyle \frac{1}{\sqrt{5}} \int \frac{du}{u\sqrt{1-(\frac{u}{\sqrt{5}})^2}}$

and then substitute

$\displaystyle sin\theta = \frac{u}{\sqrt{5}}$

$\displaystyle du= \sqrt{5}\cos\theta$

$\displaystyle u = \sqrt{5}\sin\theta$

and use the identity

$\displaystyle 1- \sin^2\theta = \cos^2\theta$

$\displaystyle \frac{1}{\sqrt{5}} \int \frac{\sqrt{5}\cos\theta}{(\sqrt{5}\sin\theta)(\co s\theta)}~d\theta = \frac{1}{\sqrt{5}} \int \csc\theta d\theta = \frac{-\ln(\csc\theta + \cot\theta)}{\sqrt{5}}$

Now just back sub

14. Originally Posted by Chop Suey
Then substitute $\displaystyle t = \frac{1}{\sqrt{5-u^2}}$ and see where that gets you.
but doing so would leave me with 1/ut right?

hm...im not sure....can you help me out..

15. Originally Posted by 11rdc11
O ok well then you can approach it like this

divide top and bottom by $\displaystyle \sqrt{5}$

which leaves

$\displaystyle \frac{1}{\sqrt{5}} \int \frac{du}{u\sqrt{1-(\frac{u}{\sqrt{5}})^2}}$

and then substitute

$\displaystyle sin\theta = \frac{u}{\sqrt{5}}$

$\displaystyle du= \sqrt{5}\cos\theta$

$\displaystyle u = \sqrt{5}\sin\theta$

and use the identity

$\displaystyle 1- \sin^2\theta = \cos^2\theta$

$\displaystyle \frac{1}{\sqrt{5}} \int \frac{\sqrt{5}\cos\theta}{(\sqrt{5}\sin\theta)(\co s\theta)}~d\theta = \frac{1}{\sqrt{5}} \int \csc\theta d\theta = \frac{-\ln(\csc\theta + \cot\theta)}{\sqrt{5}}$

Now just back sub
okay..that makes sense..thankss!