Question

a) Prove that $\stackrel{a}{\underset{a}{\int }}f\left(x\right)dx=\{\begin{array}{cc}2\stackrel{a}{\underset{0}{\int }}f\left(x\right)dx& \text{if}f\left(x\right)\text{is an even function}\\ 0& \text{if}f\left(x\right)\text{is an odd function}\end{array}$

and hence evaluate $\stackrel{1}{\underset{-1}{\int }}{\sin }^{5}x{\cos }^{4}xdx$.
b) Prove that
$\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|=1+a^2+b^2+c^2$.

Answer 1

$\left(a\right)$

${\int }_{-a}^{a}f\left(x\right)dx={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

Put $x=-t$ in first integral $⟹dx=-dt$

When $x=-a,t=a$

$x=0,t=0$

${\int }_{-a}^{a}f\left(x\right)dx=-{\int }_a^{0}f(-t)dt+{\int }_0^{a}f\left(x\right)dx$

$={\int }_0^{a}f\left(t\right)dt+{\int }_0^{a}f\left(x\right)dx$ ....... (Since, $f$ is even)

$={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$=2{\int }_0^{a}f\left(x\right)dx$

${\int }_{-a}^{a}f\left(x\right)dx=-{\int }_a^{0}f(-t)dt+{\int }_0^{a}f\left(x\right)dx$

$=-{\int }_0^{a}f\left(t\right)dt+{\int }_0^{a}f\left(x\right)dx$ ..... (Since, $f$ is odd. So, $f(-t)=-f\left(t\right)$)

$=-{\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$

$=0$

To evaluate ${\int }_{-1}^1{\sin }^{5}x{\cos }^{4}xdx$

$\sin (-x)=-\sin x$ and $\cos (-x)=\cos x$

$⟹\sin x$ is odd and $\cos x$ is even

$⟹{\sin }^{5}x$ is odd and ${\cos }^{4}x$ is even

$⟹f\left(x\right)={\sin }^{5}x{\cos }^{4}x$ is odd

$⟹{\int }_{-1}^1{\sin }^{5}x{\cos }^{4}xdx=0$ ...... ($\because f\left(x\right)={\sin }^{5}x{\cos }^{4}x$ is an odd function)

$\left(b\right)$

Consider, $\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|$

$=(a^2+1)\left|\begin{array}{cc}{b}^2+1& bc\\ cb& c^2+1\end{array}\right|-ab\left|\begin{array}{cc}ab& bc\\ ca& c^2+1\end{array}\right|+ac\left|\begin{array}{cc}ab& b^2+1\\ ca& cb\end{array}\right|$

$=(a^2+1)[b^2{c}^2+b^2+c^2+1-cbbc]-ab[ab{c}^2+ab-ab{c}^2]+ac[a{b}^{2}c-a{b}^{2}c-ac]$

$=(a^2+1)[b^2+c^2+1]-ab\left[ab\right]+ac[-ac]$

$=a^2{b}^2+a^2{c}^2+a^2+b^2+c^2+1-a^2{b}^2-a^2{c}^2$

$=a^2+b^2+c^2+1$

Hence, $\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|=1+a^2+b^2+c^2$