Question

evaluate the integral.
int t ^ { 5 } ( 2 + t ^ { 6 } ) ^ { 5 } d t
int t ^ { 5 } ( 2 + t ^ { 6 } ) ^ { 5 } d t =

Explanation:

Step1: Substitution

Let \(u = 2 + t^{6}\), then \(du=6t^{5}dt\), and \(t^{5}dt=\frac{1}{6}du\).

Step2: Change the integral limits

When \(t = 0\), \(u = 2+0^{6}=2\); when \(t = 1\), \(u = 2 + 1^{6}=3\).
The integral \(\int_{0}^{1}t^{5}(2 + t^{6})^{5}dt\) becomes \(\frac{1}{6}\int_{2}^{3}u^{5}du\).

Step3: Integrate \(u^{5}\)

Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\frac{1}{6}\times\frac{u^{6}}{6}\big|_{2}^{3}\).

Step4: Evaluate the definite integral

\(\frac{1}{36}(u^{6})\big|_{2}^{3}=\frac{1}{36}(3^{6}-2^{6})\)

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Answer:

\(\frac{665}{36}\)