# Trig. substitution

• Mar 27th 2014, 02:51 AM
kaemper
Trig. substitution
Greetings,

How is the integral (file attached) solved?

Correction: the integrand should be squared, as it equals x^2

I am aware of the pythagorian identities (cos^2x + sin^2x = 1, thus cos^2x = 1 - sin^2x). I have solved an integral with just a dx in the numerator but I have never solved an integral with an x^2 in the numerator. I don't what to do with it.

Sincere regards /

KaemperAttachment 30531
• Mar 27th 2014, 05:57 AM
romsek
Re: Trig. substitution
You didn't do it correctly. The substitution here is easy, it's the trig that gets to be a mess.

Try the substitution $x=3\sin(u)$ and be careful with the trig.
• Mar 27th 2014, 02:10 PM
kaemper
Re: Trig. substitution

I have a new integral (Attachment 30534) (this time thoroughly computed). Still it is not correct, what is wrong?
• Mar 27th 2014, 04:54 PM
romsek
Re: Trig. substitution
you haven't finished it is what isn't correct.

you are still at $\int (3 \sin(u))^2 ~du$

You can expand out the trig term and continue.
• Mar 28th 2014, 03:59 AM
kaemper
Re: Trig. substitution
It is done correctly now.