Question

evaluate the integral.
int t ^ { 8 } left( 1 + t ^ { 9 }
ight) ^ { 3 } d t
int t ^ { 8 } left( 1 + t ^ { 9 }
ight) ^ { 3 } d t =

Explanation:

Step1: Expand the integrand

Use the binomial expansion \((a + b)^3=a^{3}+3a^{2}b + 3ab^{2}+b^{3}\). Here \(a = 1\) and \(b=t^{9}\), so \((1 + t^{9})^{3}=1+3t^{9}+3t^{18}+t^{27}\). Then the integral becomes \(\int t^{8}(1 + 3t^{9}+3t^{18}+t^{27})dt=\int(t^{8}+3t^{17}+3t^{26}+t^{35})dt\).

Step2: Integrate term - by - term

Use the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).

• For \(\int t^{8}dt\), by the power rule \(\int t^{8}dt=\frac{t^{8 + 1}}{8+1}=\frac{t^{9}}{9}\).
• For \(\int3t^{17}dt\), \(\int3t^{17}dt=3\times\frac{t^{17+1}}{17 + 1}=\frac{3t^{18}}{18}=\frac{t^{18}}{6}\).
• For \(\int3t^{26}dt\), \(\int3t^{26}dt=3\times\frac{t^{26+1}}{26+1}=\frac{3t^{27}}{27}=\frac{t^{27}}{9}\).
• For \(\int t^{35}dt\), \(\int t^{35}dt=\frac{t^{35 + 1}}{35+1}=\frac{t^{36}}{36}\).

So \(\int(t^{8}+3t^{17}+3t^{26}+t^{35})dt=\frac{t^{9}}{9}+\frac{t^{18}}{6}+\frac{t^{27}}{9}+\frac{t^{36}}{36}+C\).

Answer:

\(\frac{t^{9}}{9}+\frac{t^{18}}{6}+\frac{t^{27}}{9}+\frac{t^{36}}{36}+C\)