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  • Prove the following (Exercises 34 to 39) Integral 0 to pi / 4 2 tan^3 x dx = 1 - log 2

Prove the following (Exercises 34 to 39)

    Q38.    \int_0^\frac{\pi}{4}2\tan^3 x dx = 1 - \log 2

Answers (1)

best_answer

The integral is written as

\\Let\ I=\int 2tan^{3}xdx\\ I=\int 2tan^{2}x\cdot tanxdx\\ I=\int 2(sec^{2}x-1)tanxdx\\ I=2\int tanxsec^{2}xdx-2\int tanxdx\\ I=2\int tdt-2log(cosx)+c\ \ \ \ \ \ \ (t=tanx) \\I=t^{2}-2log(cosx)+c\\ I=tan^{2}x-2log(cosx)+c

[I]_{0}^{\frac{\pi }{4}}=[tan^{2}x-2log(cosx)]_{0}^{\frac{\pi }{4}}\\

[I]_{0}^{\frac{\pi }{4}}=(1-2log\sqrt{2})-(0-2log1)

[I]_{0}^{\frac{\pi }{4}}=1-log2

Hence Proved

Posted by

Sayak

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