Question

If $f\left(x\right)=\left|\begin{array}{ccc}si{n}^{5}x& lnsinx& \frac{\sin x}{\sqrt{sinx}+\sqrt{cosx}}\\ n& {\sum }_{k=1}^{n}k& {\prod }_{k=1}^{n}k\\ \frac{8}{15}& \frac{\pi }{2}ln\left(\frac{1}{2}\right)& \frac{\pi }{4}\end{array}\right|$ Then find the value of ${\int }_0^{\pi /2}f\left(x\right)dx$

Answer 1

${\int }_0^{\pi /2}f\left(x\right)dx=\left|\begin{array}{ccc}{\int }_0^{\pi /2}si{n}^{5}xdx& {\int }_0^{\pi /2}lnsinxdx& {\int }_0^{\pi /2}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx\\ n& {\sum }_{k=1}^{n}k& {\prod }_{k=1}^{n}k\\ \frac{8}{15}& \frac{\pi }{2}ln\left(\frac{1}{2}\right)& \frac{\pi }{4}\end{array}\right|$
By Walli's formula,
${\int }_0^{\pi /2}si{n}^{5}xdx=\frac{4}{5}\cdot \frac{2}{3}=\frac{8}{15}$
Let $I={\int }_0^{\pi /2}ln\sin xdx$
$={\int }_0^{\pi /2}ln\cos xdx$
$2I={\int }_0^{\pi /2}ln\left(\frac{sin2x}{2}\right)dx$
$2I={\int }_0^{\pi /2}lnsin2xdx-{\int }_0^{\pi /2}ln2dx$
Put $2x=t$
$\Rightarrow xdx=\frac{1}{2}dt$
$2I=\frac{1}{2}{\int }_0^{\pi }ln\sin tdt-\frac{\pi }{2}ln2$
$2I=\frac{1}{2}2{\int }_0^{\pi /2}ln\sin tdt-\frac{\pi }{2}ln2$
$2I=I-\frac{\pi }{2}ln2$
$\Rightarrow I=-\frac{\pi }{2}ln2$
Let $I={\int }_0^{\pi /2}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx$
$I={\int }_0^{\pi /2}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx$
$2I={\int }_0^{\pi /2}1dx$
$\Rightarrow I=\frac{\pi }{4}$
$=\left|\begin{array}{ccc}\frac{4}{5}\cdot \frac{2}{3}& -\frac{\pi }{2}\ln 2& \frac{\pi }{4}\\ n& {\sum }_{k=1}^{n}k& {\prod }_{k=1}^{n}k\\ \frac{8}{15}& \frac{\pi }{2}\ln (\frac{1}{2})& \frac{\pi }{4}\end{array}\right|$
$=\left|\begin{array}{ccc}\frac{8}{15}& \frac{\pi }{2}ln\left(\frac{1}{2}\right)& \frac{\pi }{4}\\ n& {\sum }_{k=1}^{n}k& {\prod }_{k=1}^{n}k\\ \frac{8}{15}& \frac{\pi }{2}ln\left(\frac{1}{2}\right)& \frac{\pi }{4}\end{array}\right|$
$=0$ (since $R_1$ and $R_3$ are identical)