Question

tell me how to solve this [indefinite integral]sqrt(a+bsin^2(theta)) d(theta), a and b are here real constants.

Answer 1

\[\int\limits_{}^{}\sqrt{a+bsin^2(\theta)} d \theta\]

Answer 2

\[\int\limits_{}^{}\sqrt{a}\sqrt{1+\frac{b}{a}\sin^2(\theta)} d \theta\]

Answer 3

\[\sqrt{a}\int\limits_{}^{}\sqrt{1+\frac{b}{a} \sin^2(\theta)} d \theta \]

Answer 4

\[\sqrt{a}\int\limits_{}^{}\sqrt{1+(\sqrt{\frac{a}{b}} \sin (\theta))^2} d \theta\]

Answer 5

thinking...

Answer 6

I actually stuck in that step no subsitution or parts work here

Answer 7

maybe this might work
let \[\tan(x)=\sqrt{\frac{b}{a}}\sin(\theta)\]

Answer 8

\[\sec^2(x) dx=\sqrt{\frac{b}{a}} \cos(\theta) d \theta\]

Answer 9

nope

Answer 10

i don't think so
I appliled subsitution like that also did'nt work

Answer 11

please see my one more question. I think you can help me.

Answer 12

what?

Answer 13

i don't see another question

Answer 14

I posted but no body try to answer that question you can see in my profile

Answer 15

http://integrals.wolfram.com/index.jsp?expr=sqrt%28a%2Bb*sin+x+%5E2%29+&random=false

Answer 16

I want to know how it's solved

Answer 17

it's an elliptic integral ... good luck

Answer 18

so how it's solved