Question

Can somebody explain how to integrate sin(2t^4)t^7dt? Do you have to use u substitution?

Answer 1

@justjm do you think you know

Answer 2

is it

\(\int_{ }^{ }t^{7}\sin\left(t^{4}\right)dt\)

because then you need to do integration by parts AND u-substitution

Answer 3

Yeah, it's that integral. So what would u be set equal to in this case? Would u be equal to 2t^4 or t^4 or t^7?

Answer 4

So since the t^7 is so big...you would end up doing integration by parts 8 times!!!! but instead, you can apply u-sub first on the t^4 first, and get ride of the high degree. First do a u-sub and let u=t^4. Then you can cancel out most of the t^7 as you will notice. Then you do integration by parts

Answer 5

Hold on let me try doing it

Answer 6

\(\int_{ }^{ }t^{7}\sin\left(t^{4}\right)dt\)
let u=t^4
\(\frac{1}{4}\int_{ }^{ }t^{7}\sin\left(u\right)\frac{du}{t^{3}}\)

\(\frac{1}{4}\int_{ }^{ }t^{4}\sin\left(u\right)du\)

\(\frac{1}{4}\int_{ }^{ }u\sin\left(u\right)du\)

now use integration by parts
let u=u, so du=du
let dv=sinudu, so v=-cosu
Formula for integration by parts: \(uv-\int_{ }^{ }v'du\)
\(\frac{1}{4}\left(-u\cos u-\int_{ }^{ }-\cos udu\right)\)
\(\frac{1}{4}\left(-\cos u+\sin u+c\right)\)

YOu can plug u=t^4 back in, but all of the u's might've made it a bit confusing oof

Answer 7

Oh my bad

Answer 8

@justjm I looked at the integral incorrectly

Answer 9

It was that, but with the sin, it was actually 2*t^4

Answer 10

Oh it shouldn't be much an issue, you'd still do the same thing. Do u-sub first then the integration by parts.

Answer 11

Thank you so much, that helped a lot.