Answer :

Answer:

[tex]-\frac{1}{6} \cos (t^6) + C[/tex]

Given the indefinite integral:

[tex]\int t^t \sin(t^6) dt[/tex]

Find the integral using substitution with:

[tex]u = t^6[/tex]

First, find the derivative of u.

[tex]du = 6t^5 dt[/tex] (power rule)

Then, solve for dt so that we can substitute it back into the integral.

[tex]\frac{du}{6t^5} = dt[/tex]

Substitute.

[tex]\int t^5 \sin(t^6) \times \frac{du}{6t^5}[/tex]


The terms: [tex]t^5[/tex] cancel out.

So it becomes:

[tex]\int \frac{1}{6} \sin(u) du[/tex]

Integrate.

[tex]\frac{1}{6} \int \sin(u) du \\--- > \frac{1}{6} \times -\cos(u)[/tex]

Substitute back in u.

[tex]Integral = -\frac{1}{6} \cos(t^6) + C[/tex]

Don't forget plus constant C as this is an indefinite integral!