Question

a) Prove that $\stackrel{a}{\underset{0}{\int }}f\left(x\right)dx=\stackrel{a}{\underset{0}{\int }}f(a-x).dx$ and hence evaluate
$\stackrel{\pi /4}{\underset{0}{\int }}\log (1+\tan x)dx$
b) Find the value of $k$, if
$f\left(x\right)=\{\begin{array}{cc}\frac{k\cos x}{\pi -2x}& ifx\ne \frac{\pi }{2}\\ 3& ifx=\frac{\pi }{2}\end{array}$
is continuous at $x=\frac{\pi }{2}$.

Answer 1

$\left(a\right)$

Consider ${\int }_0^{a}f\left(x\right)dx$ ........... $\left(1\right)$

Put $a-x=u⟹-dx=du$

When $x=0⟹u=a$

When $x=a⟹u=0$

$⟹{\int }_0^{a}f\left(x\right)dx=-{\int }_a^{0}f(a-u)du$

$={\int }_0^{a}f(a-u)du$

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

$\therefore I={\int }_0^{\pi /4}\log (1+\tan x)dx={\int }_0^{\pi /4}\log (1+\tan [\frac{\pi }{4}-x)])dx$ .... Using above inequality

$={\int }_0^{\pi /4}\log (1+\frac{\tan \left(\frac{\pi }{4}\right)-\tan x}{1+\tan \left(\frac{\pi }{4}\right)\tan x})$

$={\int }_0^{\pi /4}\log (1+\frac{1-\tan x}{1+\tan x})$

$={\int }_0^{\pi /4}\log \left(\frac{2}{1+\tan x}\right)$

$={\int }_0^{\pi /4}\log 2-\log (1+\tan x)dx$

$={\int }_0^{\pi /4}\log 2-{\int }_0^{\pi /4}\log (1+\tan x)dx$

$={\int }_0^{\pi /4}\log 2-I$

$⟹2I=\log 2[x{]}_0^{\pi /4}$

$⟹2I=\frac{\left(\log 2\right)\pi }{4}$

$⟹I=\frac{\pi }{8}\log 2$

$\left(b\right)$

Given $f\left(x\right)=\{\begin{array}{cc}\frac{k\cos x}{\pi -2x}& ifx\ne \frac{\pi }{2}\\ 3& ifx=\frac{\pi }{2}\end{array}$

is continuous at $x=\frac{\pi }{2}$

$⟹{lim}_{x\to \frac{\pi }{2}}f\left(x\right)=f\left(\frac{\pi }{2}\right)$

$⟹{lim}_{x\to \frac{\pi }{2}}\frac{k\cos x}{\pi -2x}=3$

$⟹{lim}_{x\to \frac{\pi }{2}}\frac{-k\sin x}{-2}=3$ ..... Using L'Hospital's rule

$⟹\frac{-k\sin \left(\frac{\pi }{2}\right)}{-2}=3$

$⟹-k=-6$

Hence, $k=3$