Prove the following Exercise: ^ 2 _ 0 2 ^ 3 x dx=1- 2 - Vidyadip

Question

Prove the following Exercise:
$\int^{\frac{\pi}{2}}_{0}2\tan^{3}\text{x dx}=1-\log2$

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Answer

$\text{Let I}=\int^{\frac{\pi}{4}}\limits_{0}2\tan^{3}\text{x dx}$
$\text{I}=\int^{\frac{\pi}{4}}\limits_{0}\tan^{2}\text{x}\ \tan\text{x dx}=2\int^{\frac{\pi}{4}}\limits_{0}(\sec^{2}\text{x}-1)\tan\text{x dx}$
$=2\int^{\frac{\pi}{4}}\limits_{0}\sec^{2}\text{x}\ \tan\text{x dx}-2\int^{\frac{\pi}{4}}\limits_{0}\tan\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_0\sec\text{x}.\sec\text{x}\tan\text{x dx}-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$\text{Now,} \sec\text{x}=\text{t}$
$=2\Big[\frac{\text{t}^2}{2}\Big]^{\sqrt{2}}_1-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$=2\Big[\frac{2}{2}-\frac{1}{2}\Big]-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$=1-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
Hence, the given result is Proved

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