Question

evaluate the integral.
int t ^ { 2 } left( 5 + t ^ { 3 }
ight) ^ { 8 } d t
int t ^ { 2 } left( 5 + t ^ { 3 }
ight) ^ { 8 } d t = square

Explanation:

Step1: Use substitution

Let \(u = 5 + t^{3}\), then \(du=3t^{2}dt\), and \(t^{2}dt=\frac{1}{3}du\).

Step2: Rewrite the integral

The integral \(\int t^{2}(5 + t^{3})^{8}dt\) becomes \(\frac{1}{3}\int u^{8}du\).

Step3: Integrate \(u\)

Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have \(\frac{1}{3}\times\frac{u^{9}}{9}+C=\frac{u^{9}}{27}+C\).

Step4: Substitute back \(u\)

Substituting \(u = 5 + t^{3}\) back, we get \(\frac{(5 + t^{3})^{9}}{27}+C\).

Answer:

\(\frac{(5 + t^{3})^{9}}{27}+C\)