Anyone good at calc?

Question

I've had these up but no one has been able to help me with the following. The first of the three seems to be the trickiest...

Use Integration by parts to evaluate the integrals:

[IntSign](ln(4x))^2 dx

[IntSign]x^(9)cos(x^5) dx

Evaluate the following:

[IntSgn]sin^3(4x)cos^8(4x) dx

Answer 1

∫ln² (4x) dx

The integral of ln isn't obvious - its derivative is. Therefore, we want to make u=ln² (4x), which means se have to let dv=dx. Then v=x and du=2 ln (4x) *1/(4x) * 4 dx=2 ln (4x)/x dx. So integrating by parts we have:

x ln² (4x) - ∫2 x ln (4x)/x dx
x ln² (4x) - 2 ∫ln (4x) dx

We use the same trick again: u=ln (4x), du=4/(4x) dx = 1/x dx, v=x, dv=dx. Then:

x ln² (4x) - 2 (x ln (4x) - ∫x/x dx)
x ln² (4x) - 2x ln (4x) + 2∫1 dx
x ln² (4x) - 2x ln (4x) + 2x + C

For the next one, we need to use u-substitution before we can do anything. Let u=x⁵, then du=5x⁴ dx, so x⁴ dx = 1/5 du, so

∫x⁹ cos (x⁵) dx
∫x⁵ cos (x⁵) x⁴ dx
1/5 ∫u cos u du

At this point, we begin using integration by parts. Let v=u, dv=du, w=sin u, dw=cos u du. So we have:

1/5 u sin u - 1/5 ∫sin u du
1/5 u sin u + 1/5 cos u + C
x⁵/5 sin (x⁵) + 1/5 cos (x⁵) + C

For the last one, remember your trig identities:

∫sin³ (4x) cos⁸ (4x) dx
∫cos⁸ (4x) (1-cos² (4x)) sin (4x) dx
∫cos⁸ (4x) sin (4x) - cos¹⁰ (4x) sin (4x) dx

Let u=cos (4x), du=-4 sin (4x) dx, so we have:

-1/4 ∫u⁸ - u¹⁰ du
u¹¹/44 - u⁹/36 + C
cos¹¹ (4x)/44 - cos⁹ (4x)/36 + C

And we are done.

Answer 2

If i was good at calc I wouldn't have time to do this. I would have a real job making real$$$.