1. ## Integrate sin^2(4t) cos(4t)

Hi,

I have the follwing that I am struggling with:

\int sin^2(4t) cos(4t).dt

so far I have got:

u = sin^2(4t)

du/dt = 4t

du = 1/4

Now I am not sure what to do with the cos(4t) can someone help explain and show me the answewer so I maybe able to work backwards?

Thanks

m

2. Originally Posted by mm874
Hi,

I have the follwing that I am struggling with:

\int sin^2(4t) cos(4t).dt

so far I have got:

u = sin^2(4t)

du/dt = 4t

du = 1/4

Now I am not sure what to do with the cos(4t) can someone help explain and show me the answewer so I maybe able to work backwards?

Thanks

m
We had this discussion only recently. It is entirely unnecessary to use a u-substitution. You can just apply the general rule that:

$\int f'(x)[f(x)]^n dx=\dfrac{[f(x)]^{n+1}}{n+1}+C$ or you can just use the fact that this clearly resembles the derivative of $sin^3(4t)$.

If you absolutely have to use substitution (and you really don't) then your $\frac{du}{dt}$ is incorrect. How do you differentiate $Sin^2(4t)$?

3. The substitution you are supposed to make is $\displaystyle u = \sin{(4t)} \implies du = 4\cos{(4t)}\,dt$.

4. Originally Posted by Prove It
The substitution you are supposed to make is $\displaystyle u = \sin{(4t)} \implies du = 4\cos{(4t)}\,dt$.
I agree that this is better, although the $sin^2(x)$ substitution worked when I tried it just now, although it wasn't simple.