Integrate:Sinxcosx/a^2cos^2(x)+b^2sin^2(x))dx???

Question

plz i need the full working too hard for me

Answer 1

∫(sinx cosx)/(a^2 cos^2(x) + b^2 sin^2(x)) dx

well, let's see.

a^2 cos^2(x) + b^2 sin^2(x)
= a^2 cos^2(x) + a^2 sin^2(x) + (b^2-a^2)sin^2(x)
= a^2 + (b^2-a^2) sin^2(x)

so, if
u = a^2 + (b^2-a^2) sin^2(x)
du = 2(b^2-a^2) sinx cosx dx

and your integral now becomes

1/2 ∫ du/u

Not so hard now, eh?

Or, I guess you could just let

u = a^2 cos^2(x) + b^2 sin^2(x)
du = -2a^2 sinx cosx + 2b^2 sinx cosx
= 2sinx cosx (b^2-a^2)
and you have
1/(2(b^2-a^2)) ∫ du/u