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NCERT Class XII Mathematics Chapter - - Solutions

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Prove the following (Exercises 262 to 267)
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Question : 266 of 271
Marks: +1, -0
∫0π/42tan⁡3xdx=1−log⁡2\int\limits_{0}^{\pi/4}2\tan^3 x dx=1-\log 2
Solution:  
∫0π/42tan⁡3xdx\int\limits_{0}^{\pi/4}2\tan^3 x dx
=∫0π/42tan⁡x⋅tan⁡xdx=\int\limits_{0}^{\pi/4}2\tan x\cdot\tan^x dx
=∫0π/42tan⁡x(sec⁡2x−1)dx=\int\limits_{0}^{\pi/4}2\tan x(\sec^2 x-1)dx
=2∫0π/4(tan⁡x)sec⁡2xdx−2∫0π/4tan⁡xdx=2\int\limits_{0}^{\pi/4}(\tan x)\sec^2 x dx-2\int\limits_{0}^{\pi/4}\tan x dx
=2[tan⁡2x2]0π/4−2[−log⁡∣cos⁡x∣]0π/4=2\left[\frac{\tan^2 x}{2}\right]^{\pi/4}_{0}-2\left[-\log|\cos x|\right]^{\pi/4}_{0}
=(tan⁡2π4−tan⁡20)+2(log⁡cos⁡π4−log⁡cos⁡0)=\left(\tan^2\frac{\pi}{4}-\tan^2 0\right)+2\left(\log\cos\frac{\pi}{4}-\log\cos 0\right)
=(1−0)+2(log⁡(1/2)−log⁡1)=(1-0)+2\left(\log(1/\sqrt{2})-\log 1\right)
=1+2(log⁡1−log⁡2−log⁡1)=1+2(\log 1-\log\sqrt{2}-\log 1)
=1+2×(−12log⁡2)=1+2\times\left(-\frac{1}{2}\log 2\right)
=1−log⁡2=1-\log 2
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