Question

${\int }_{-1}^1\frac{9x^5+4x^3+2x+1}{x^2+4}dx=k$ then $5\sin k$ lies in the interval.

Answer 1

$\begin{array}{c}I={\int }_{-1}^1\frac{9x^5+4x^3+2x+1}{x^2+4}dx\\ I={\int }_{-1}^1\frac{-9x^5+4x^3-2x+1}{x^2+4}dx\\ 2I={\int }_{-1}^1\frac{2}{x^2+4}dx\\ I={\int }_{-1}^1\frac{dx}{x^2+4}\\ I=2{\int }_0^1\frac{dx}{x^2+4}\\ I=2\times \frac{1}{2}{\left[{\tan }^{-1}\left(\frac{2}{2}\right)\right]}_0^1\\ I={\tan }^{-1}\left(\frac{1}{2}\right)\\ k={\tan }^{-1}\left(\frac{1}{2}\right)\\ 5\sin \left({\tan }^{-1}\left(\frac{1}{2}\right)\right)=5\sin {\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)=\sqrt{5}\\ 2<\sqrt{5}<3\end{array}$