# Thread: Solving an integral using a special substitution

1. ## Solving an integral using a special substitution

Can somebody tell me if this is right?

I must solve:
$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dx}{\sqr t[ ]{9+x^2}}$

And I used the substitutions:
$\displaystyle x=3sinht$
$\displaystyle dx=3coshtdt$

And the identity:

$\displaystyle cosh^2t-sinh^2t=1\Rightarrow{\cosht=\sqrt[ ]{1+sinh^2t}}$

$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dx}{\sqr t[ ]{9+x^2}}=\displaystyle\int_{}^{}\displaystyle\frac {3coshtdt}{\sqrt[ ]{3^2+(3sinht)^2}}=\displaystyle\int_{}^{}\displays tyle\frac{3coshtdt}{\sqrt[ ]{3^2(1+sinh^2t)}}=\displaystyle\int_{}^{}\displays tyle\frac{coshtdt}{cosht}=\displaystyle\int_{}^{}d t=t+C=argsh(\displaystyle\frac{x}{3})+C$

Bye there, and thanks for posting.

2. Originally Posted by Ulysses
Can somebody tell me if this is right?
Well done!

3. Thanks.

$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin (t)}$

I've solved this using the table, using the definition of cosecant: $\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin (t)}=\ln|csc(t)-cot(t)|+C$

This was the result of solving: $\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dx}{x\sq rt[ ]{1-x^2}}=\ln|csc(arcsin(x))-cot(arcsin(x))|+C$

But now I must solve:
$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\ sqrt[ ]{4-x^2}}$

$\displaystyle x=2\sin(t)$
$\displaystyle dx=2\cos(t)dt$

$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\ sqrt[ ]{4-x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{2\ cos(t)dt}{4\sin^2(t)\sqrt[ ]{2^2-(2\sin(t)^2}}=\displaystyle\frac{1}{4}\displaystyl e\int_{}^{}\displaystyle\frac{dt}{\sin^2(t)}}=\dis playstyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t )dt$

I thought of solving by parts, but...

$\displaystyle u=csc(t)$
$\displaystyle du=-csc(t)cot(t)dt$

$\displaystyle dv=csc(t)dt$
$\displaystyle v=\ln|csc(t)-cot(t)|$

4. Originally Posted by Ulysses

$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin (t)}$
Do you know the Weierstrass substitution? If If $\displaystyle \displaystyle \tan\dfrac{t}{2} = x$, then $\displaystyle \displaystyle \sin\dfrac{t}{2} = \dfrac{2x}{1+x^2}$ and $\displaystyle dt = \dfrac{2}{1+x^2}\;{dx}$. Using these results we have $\displaystyle \displaystyle \int\dfrac{dt}{\sin{t}}\;{dt} = \int\bigg(\dfrac{1}{\frac{2x}{1+x^2}}\bigg)\bigg(\ dfrac{2}{1+x^2}\bigg)\;{dx} = \int\dfrac{1}{x}\;{dx} = \ln{x}+k = \ln\left(\tan\dfrac{t}{2}\right)+k$

But now I must solve:
$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\ sqrt[ ]{4-x^2}}$

$\displaystyle x=2\sin(t)$
$\displaystyle dx=2\cos(t)dt$

$\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\ sqrt[ ]{4-x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{2\ cos(t)dt}{4\sin^2(t)\sqrt[ ]{2^2-(2\sin(t)^2}}=\displaystyle\frac{1}{4}\displaystyl e\int_{}^{}\displaystyle\frac{dt}{\sin^2(t)}}=\dis playstyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t )dt$
$\displaystyle \displaystyle\int\dfrac{1}{\sin^2{x}}\;{dx} = -\cot{x}+k$ is a standard integral. As long you broke down the integral up-to this stage and know this result you should be alright.

5. Thank you verymuch!

6. Originally Posted by Ulysses
Thank you verymuch!
No problem. I take a delight in helping people who show their work, if I can that is.

7. Really noble from you sharing your knowledge that way. Believe me, I know that people like you makes this world a better place. I know... a post in a forum doesn't mean much, but is the attitude. And actually the free access to knowledge makes the difference.

Thanks is all I can give now