Question

Evaluate the integral
\[\int_0^{\pi/2} \frac{\cos t}{\sqrt{1+\sin^2 t}} \text{d}t\]

Answer 1

\[\int \frac{\cos t}{\sqrt{1+\sin^2 t}} \text{d}t\]

Answer 2

^that's what I assumed...

Answer 3

\[\int \frac{\cos t}{\sqrt{2-\cos^2 t}} \text{d}t\]emm what do u want to do with this?

Answer 4

oh dang, this time I wasn't thinking :P

Answer 5

one of those days... I'll be back for this one I guess

Answer 6

You could use the same thing but with hyperbolic functions, substitute sin t by sinh t then 1+sinh^2(t) by cosh^2(t)

Answer 7

aha!\[\sin x=\tan\theta\implies \cos xdx=\sec^2\theta d\theta\]\[\int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}d\theta=\int \sec\theta d\theta=\ln|\sec\theta+\tan\theta|\]undo the substitutions\[\sec\theta=\sqrt{1+\sin^2x}\]\[\tan\theta=\sin x\]\[=\ln|\left.\sqrt{1+\sin^2x}+\sin x|\right|_0^{\pi/2}=\ln|\sqrt2+1|\]stil don't know how to get a hyperbolic sine for that, but hey, I'm happy :)

Answer 8

I trust your method too @ivanmlerner but I never made a sub with a hyperbolic trig function so I wanted another way ;)