evaluate the integral, wee! u-sub

[voidflux]

Hello everyone,
I"m having troubles finding what expression I should u-sub. here's the problem.

definate integral, from 0 to pi/2   [cos(x)/(1+sin^2(x))]dx

I tried
u = 1+sinx;
du = cosx dx;

but that won't due because the sin^2(x),
like i can't do this:
integral symbol du/u^2
because (1+sin^2 x ) != (1+sinx)^2 or is it?

also is this true or false? 2
definate integral from -1 to 2 1/x dx = [ln|x|] = ln2 - ln1 = ln2
-1

Thanks!

[TheManWhoCan]

Integration by parts, if you already know how read this:

S{cosx/(1+sin²x)}dx = S(1/(1+sin²x)}d(sinx) = tan^-1(sinx)

If you don't know about integration by parts, substitute u = sinx (it amounts to the same thing):

u = sinx: du = cosx.dx

You'll reduce the integral to:

S{1/(1 + u²)}du = tan^-1u

You changed the variable so the limits become 0 and 1:

[tan^-1u]{0,1} = tan^-11 = pi/4

And:

S{1/x}dx = ln|x|

[ln|x|]{-1,2} = ln2 - ln1 = ln2

[voidflux]

Thanks for the responce!