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Question

Question: \[\int_{0}^{\pi/4}{\tan^{2}xdx =}\]...

∫0π/4tan⁡2xdx=\int_{0}^{\pi/4}{\tan^{2}xdx =}

A

1−π41 - \frac{\pi}{4}

B

1+π41 + \frac{\pi}{4}

C

π4−1\frac{\pi}{4} - 1

D

π4\frac{\pi}{4}

Answer

1−π41 - \frac{\pi}{4}

Explanation

Solution

∫0π/4tan⁡2xdx=∫0π/4(sec⁡2x−1)dx\int_{0}^{\pi/4}{\tan^{2}xdx = \int_{0}^{\pi/4}{(\sec^{2}x - 1)dx}}

=∫0π/4sec⁡2xdx−∫0π/41dx= \int_{0}^{\pi/4}{\sec^{2}xdx - \int_{0}^{\pi/4}{1dx}}= [tan⁡x]0π/4−[x]0π/4=1−π4\lbrack\tan x\rbrack_{0}^{\pi/4} - \lbrack x\rbrack_{0}^{\pi/4} = 1 - \frac{\pi}{4}.