- #1
- 493
- 8
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?
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?