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Prove the following\[\int\limits_0^\frac{\pi}{4}2\tan^3x\;dx=1-log2\]

1 Answer

  • (i)$\int \limits _a^b f(x)dx=F(b)-F(a)$
  • (ii) $ \tan ^2 x=\sec ^2 x-1$
  • (iii)$ \int \tan x dx=-log |\cos x| +c$
  • To prove $\int \limits_0^{\pi/4} 2\tan ^3x dx=1-log 2$
Given $ I=\int \limits_0^{\pi/4} 2 \tan ^3 x dx$
Let us write $ \tan ^3 x \;as\;\tan^2x.\tan x$
Hence $I=2 \int \limits_0^{\pi/4} \tan ^2x.\tan x dx$
But $\tan ^2x=\sec ^2 x-1$
Therefore $I=2 \int \limits_0 ^{\pi/4} (\sec ^2x-1)\tan x dx$
On seperating the terms,
$I=2[\int \limits_0^{\pi/4} \sec ^2 x \tan x]dx-2[\int \limits_0^{\pi/4}\tan x dx]$
on integrating ,
Let $I=2[I_1-I_2]$
Consider $I_1=\int \limits_0^{\pi/4} \tan x \sec^2x dx$
Let $\tan x =t,$ on differentiating w.r.t.x we get $\sec ^2xdx=dt$
As we substitute t , the limits also changes,
when x=0,since $\tan x=0 \qquad t=0$
when $x=\pi/4,$since $\tan \pi/4=1 \qquad t=1$
Therefore $I_1=\int \limits_0^1 t.dt$
Hence $ I=2 \bigg[\int \limits_0^1 t.dt- \int \limits_0^{\pi/4} \tan x dx\bigg]$
on integrating we get,
$I=2 \bigg\{\bigg [\frac{t^2}{2}\bigg]_0^1-\bigg[-log \cos x \bigg]_0^{\pi/4}\bigg\}$
On applying limits,
$I=2\bigg\{\bigg[(\frac{1}{2}\bigg)-0\bigg]+\bigg[log \cos \pi/4-log \cos 0 \bigg]\bigg\}$
But $ \cos \pi/4=\frac{1}{\sqrt 2}\; and \; \cos 0=1$
$=2 \times \frac{1}{2}+2 [ log \frac{1}{2\sqrt 2}-log 1]$
$=1+\log (\frac{1}{\sqrt 2})^2-0 \qquad(log 1=0)$
$=1+\log (1/2)$
$=1-log 2$
Hence proved



answered Mar 14, 2013 by meena.p