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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Prove $\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2$
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Answer

Given integral is: $\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x$ 
To Prove: $\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2$ 
Let $I=\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x  ...(i)$
= $\int_{0}^{\frac{\pi}{4}} 2 \cdot \tan x \cdot \tan ^{2} x d x$ 
= $\text { 2. } \int_{0}^{\frac{\pi}{4}} \tan x \cdot\left(\sec ^{2} x-1\right) d x$ 
$\Rightarrow \mathrm{I}=2\left\{-\int_{0}^{\frac{\pi}{4}} \tan \mathrm{x} \mathrm{dx}+\int_{0}^{\frac{\pi}{4}} \tan \mathrm{x} \cdot \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}\right\}$ 
$\Rightarrow \mathrm{I}=-[2 \log \cos \mathrm{x}]_{0}^{\pi / 4}+2 . \mathrm{I}_{1} ...(ii)$
Solving $I_1:$
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{4}} \tan \mathrm{x} \cdot \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}$ 
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{4}} \tan \mathrm{x} \cdot \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}$ 
Let, tan $x = t $
$\Rightarrow \sec^2 x dx = dt$
When $x = 0$ then $t = 0$ and when $x = \frac{\pi}{4}$ then $t = 1$
$\Rightarrow \mathrm{I}_{1}=\int_{0}^{1} \mathrm{t} \mathrm{dt}$ 
$= \left[\frac{t^{2}}{2}\right]_{0}^{1}$ 
$\Rightarrow \mathrm{I}_{1}=\frac{1}{2}$ 
Using this in equation $(ii)$
$\Rightarrow \mathrm{I}=[2 \log \cos \mathrm{x}]_{0}^{\pi / 4}+2 \cdot \frac{1}{2}$ 
$\Rightarrow \mathrm{I}=2\left\{\log \cos \frac{\pi}{4}-\log \cos 0\right\}+1$ 
$\Rightarrow \mathrm{I}=2\left\{\log \frac{1}{\sqrt{2}}-\log 1\right\}+1$ 
$\Rightarrow \mathrm{I}=\left\{\log \left(\frac{1}{\sqrt{2}}\right)^{2}-\log (1)^{2}\right\}+1$ 
$\Rightarrow \mathrm{I}=1-\log 2+\log 1$ 
$\Rightarrow I=1-\log 2$ 
Hence Proved.

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