
1 Attachment(s)
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

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.

1 Attachment(s)
Re: Trig. substitution
Thank you this answer.
I have a new integral (Attachment 30534) (this time thoroughly computed). Still it is not correct, what is wrong?

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.

Re: Trig. substitution
It is done correctly now.