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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!