Integrate sinx cosx dx: Find Solution

[Natasha1]

I have been asked to find the integral sinx cox dx using the identity sin2x = 2sinxcosx

My work...

integral of sinx cox dx

= 1/2 integral of 2 sinx cos dx

= 1/2 integral of sin 2x dx

u = 2x
du = 2

so 1/2 * 1/2 of integral of sin u du

= 1/4 [-cos u] + c
= - 1/4 cos 2x + c is this correct?

 

[arildno]

Quite so!

 

[0rthodontist]

Yes, it is, except you should say du = 2 dx instead of just 2.

 

[Natasha1]

Yes, it is, except you should say du = 2 dx instead of just 2.

Great. So if we know look at the following integral x sin x cos x dx

My work...

Let u = x

du/dx = 1

dv/dx = sin x cos x

v = -1/4 cos 2x (from above)

so = -1/4 x cos x - integral of -1/4 cos 2x dx

= -1/4 x cos 2x + 1/4 integral of cos 2x dx

Let u = 2x

du/dx = 2

so = -1/4 x cos 2x + 1/4 * 1/2 integral of cos u du

= -1/4 x cos 2x + 1/8 sin 2x + c is this correct?

 

[arildno]

Yes it is, as can be verified by differentiating your expression.