Integral help!?

Question

What is the integral of 4θ+cos8θ ?

That's theta, not 0.

I'd appreciate any help. Thanks

Answer 1

I = ∫ 4Ɵ dƟ + ∫ cos 8θ dθ
I = 2 θ ² + (1/8) sin 8θ + C

Answer 2

I = ∫(4θ+cos8θ)dθ
I = 4∫θdθ + ∫cos8θdθ
I = 4(1/2)θ² + (1/8)sin8θ + C (for arbitrary constant C)
I = 2θ² + (1/8)sin8θ + C

Answer 3

int( 4θ )dθ + int( cos(8θ) )dθ

4 · θ^2 / 2 + (sin(8θ)) / 8 + c
[reverse power rule; simple substitution]

2θ^2 + (sin(8θ)) / 8 + c

Because an integral or anti derivative is the reverse operation we can check our answer by taking the derivative of our result. Sure enough:

f(x) = 2θ^2 + (sin(8θ)) / 8 + c
f'(x) = 2 ·2 · θ^(2 - 1) + 8 · cos(8θ) / 8 + 0
f'(x) = 4θ + cos(8θ)

If you need any extra help, feel free to email me.

Answer 4

int[4θ+cos8θ ]=
int[4θ]+int[cos8θ]=
2 θ^2+ 1/8*sin8θ