integral qz plz help? Ssinx2dxcos2x=?

Answer 1

I'm trying to guess what integrand you have here. My best guess is that you want the integral of
(sin x)^2 dx / cos 2x

Now (sin x)^2 = (1 - cos 2x)/2, and so we want integral of

1/(2cos(2x)) - 1/2 with respect to x, which is

(1/4) ln(tan 2x + sec 2x) - x/2, if my recollection of standard integrals is correct.

+ C, of course.

Answer 2

So you want to solve

Integral (sin^2(x) cos^2(x)) dx

Keep in mind that

sin^2(x) cos^2(x) = [sin(x)cos(x)]^2

Also note that sin2x = 2sinxcosx. That means
(1/2)sin(2x) = sin(x)cos(x)

Therefore

[sin(x)cos(x)]^2 = [(1/2)sin(2x)]^2 = (1/4)sin^2(2x)

That means we want to take the integral of that.

Integral [ (1/4) sin^2(2x)] dx

It's always a good idea to pull out constants, so let's pull out the 1/4.

1/4 * Integral [sin^2(2x)]dx

Now, recall the half angle identity, which goes as follows.

sin^2(y) = (1/2)[1 - cos2y]

In our case, we have 2x; therefore

sin^2(2x) = (1/2) [1 - cos(2(2x))]
sin^2(2x) = (1/2) [1 - cos(4x)]

So now we replace what we currently have as our integral with that.

1/4 * Integral [sin^2(2x)]dx =

(1/4) * Integral ( (1/2) [1 - cos(4x)] )dx

Pull out the constant again; the 1/2 will merge with the 1/4 to make 1/8.

(1/8) * Integral (1 - cos(4x))dx

Now, the integral is trivial, and we get

(1/8) * [ x - (1/4) sin(4x) ] + C

(1/8)x - (1/32) sin(4x) + C

Answer 3

Please be more clear as to what the integrand really is...the present information is confusing.