Simplifying a Trigonometric Integral

[adoado]

This is not a homework question. I am doing extra work to get a grasp on integration and I am not sure if this is correct...

$$\int(sin^2x*cos^2x)dx$$

I did this:

1) $$\int(sinx*cosx)*(sinx*cosx)dx$$

2) $$\int(1 /2 sin(2x)) * (1 /2 sin(2x)) dx$$ (Double angle for sine formula)

3) $$(1/4) \int(sin^2(2x))dx$$

4) $$(1/4) * (1/2) \int(1-cos2x)dx$$ (Reduction of square power)

5) = 1/8[x + (1/2)sin2x] + C

I am not sure if this is correct, just hoping someone could help me out a little or confirm it ^^

Cheers,
Adrian

 

[adriank]

Almost right, except that in going from step 3 to 4, you should have instead
$$(1/4) * (1/2) \int(1 - \cos 4x) \,dx.$$
You also have the sign on the sin term wrong in going from step 4 to 5 (the antiderivative of cos is sin, not -sin). The answer would then be
$$\frac18 \left( x - \frac14 \sin 4x \right).$$

 

[adoado]

Thanks adriank, its good to know I was at least partly there ^^

Cheers,
Adrian