Question

Can you plase help me on this ..

Question is Evaluate the integral using power series .
integral of sinh x^3 dx = limit 0 to 0.5

Answer 1

\[\begin{align*}\sinh x&=\frac{e^{x}-e^{-x}}{2}\\\\
&=\frac{1}{2}\left(\sum_{n=0}^\infty\frac{x^n}{n!}-\sum_{n=0}^\infty\frac{(-x)^n}{n!}\right)\\\\
&=\frac{1}{2}\left[\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\right)-\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots\right)\right]\\\\
&=x+\frac{x^3}{6}+\frac{x^5}{120}+\cdots\\\\
&=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}
\end{align*}\]
This means
\[\sinh x^3=\sum_{n=0}^\infty\frac{(x^3)^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty\frac{x^{6n+3}}{(2n+1)!}\]
Compute the integral:
\[\int_0^{1/2}\sum_{n=0}^\infty\frac{x^{6n+3}}{(2n+1)!}~dx\]