Question

Evaluate the integral

Integral of cost/(1+(sint)^2)^1/2 from 0 to pi over 2

Answer 1

seeing that cos is the derivative of sin; youd think there was a chain rule involved

Answer 2

\[\int\frac{cos(t)}{(1+sin^2(t))^{1/2}}dt \]
\[\int cos(t)(1+sin^2(t))^{-1/2}dt \]
what happens when we try to derive something close to it?
\[\frac{d}{dx}\ (1+sin^2(t))^{1/2}\]
\[\frac{(1+sin^2(t))'}{2(1+sin^2(t))^{1/2}}\]
\[\frac{2\ sin(t)\ cos(t)}{2(1+sin^2(t))^{1/2}}\]
\[\frac{sin(t)\ cos(t)}{(1+sin^2(t))^{1/2}}\]
which is close, but is it helpful?

Answer 3

Substitute u = sin t
And you'll get the integral of 1/sqrt(1+u^2)

Now make a second substitution, u = sinh t (that's hyperbolic sin) and you'll get the integral of 1 dt

Answer 4

my other thought is to use an ln(...) config

Answer 5

doh!! ... hyperbolics

Answer 6

Thank you, this has all been very helpful.