evaluate the integral.\n int t ^ { 2 } left( 5 + t ^ { 3 } \right) ^ { 8 } d t \n int t ^ { 2 } left( 5 + t…

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

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

Answer

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\neq - 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)