Browse Questions

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

Toolbox:
• (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