find the indefinite integrals of h(u)=cos^2(5u)?

Answer 1

Integral cos² 5u du
=
1/2 Integral (1 + cos 10u) du
=
1/2 [u + (sin 10u)/10]

Answer 2

Since cos 2x = 2cos^2 x -1
So cos^2 x = 1/2(1+cos 2x)
Int cos^2(5u) du = int [1/2 (1+cos (10u))] du
= 1/2 u + 1/20 sin (10u) + c

Answer 3

I = ∫ cos² 5u.du
Let x= 5u
dx = 5 du and dx/5 = du
I = (1/5).∫cos² x dx
I = (1/5).∫(1/2).cos 2x - 1/2 .dx
I = (1/20).sin 2x - (1/10).x + c
I = (1/20).sin 10u - (1/2).(u) + c

Answer 4

int(cos)^2(5u)du
let t=5u;du=dt/5;
int(cos)^2(5u)du
cos^2(t)=1/2(cos(2t)+1)
=(1/5)(1/2)int(cos(2t)+1)dt
=(1/5)(1/4)(sin(2t)+t/2)+C
but,t=5u
therefore,
int(cos)^2(5u)du
=(1/20)[10u+sin(10u)]+C

i hope that this helps