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Integrals — Maths STD 12 Science — Question

Gujarat 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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