Question

Evaluate the integral.
∫(from π/6 to 3π/4) π(sin^2x+2)^2 - 4

Answer 1

∫[π/6,3π/4] π*(sin^2x+2)^2 - 4 dx
I guess you're having trouble with the integration?
∫(sin^2x + 2)^2 dx
= ∫sin^4x + 4sin^2x + 4 dx
Now break that up into
∫ sin^2x dx = ∫ (1-cos2x)/2 dx = 1/2 ∫ 1 - cos2x dx
which is easy
∫sin^4 dx = ∫((1-cos2x)/2)^2 dx
= 1/4 ∫ 1 - 2cos2x + cos^2 2x dx
which is all easy except
∫cos^2 2x dx = (1+cos4x)/2 dx = 1/2 ∫ 1 + cos4x dx
so now put all those pieces together

you can check your result an any of several handy integral calculators online.

wolframalpha.c om is one