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Question

Question: \[\int_{}^{}{\frac{\tan x}{\sqrt{1 - \tan^{2}x}}dx =}\]...

∫tan⁡x1−tan⁡2xdx=\int_{}^{}{\frac{\tan x}{\sqrt{1 - \tan^{2}x}}dx =}

A

cos⁡h−1(2cos⁡x)+c\cos h^{- 1}(\sqrt{2}\cos x) + c

B

−cos⁡h−1(2cos⁡x)+c- \cos h^{- 1}(\sqrt{2}\cos x) + c

C

12cos⁡−1(2cos⁡x)+c\frac{1}{\sqrt{2}}\cos^{- 1}(\sqrt{2}\cos x) + c

D

−12cos⁡h−1(2cos⁡x)+c\frac{- 1}{\sqrt{2}}\cos h^{- 1}(\sqrt{2}\cos x) + c

Answer

−12cos⁡h−1(2cos⁡x)+c\frac{- 1}{\sqrt{2}}\cos h^{- 1}(\sqrt{2}\cos x) + c

Explanation

Solution

∫tan⁡xdx1−tan⁡2x=∫sin⁡xdxcos⁡2x−sin⁡2x=∫sin⁡xdx2cos⁡2x−1\int_{}^{}{\frac{\tan xdx}{\sqrt{1 - \tan^{2}x}} = \int_{}^{}{\frac{\sin xdx}{\sqrt{\cos^{2}x - \sin^{2}x}} = \int_{}^{}\frac{\sin xdx}{\sqrt{2\cos^{2}x - 1}}}}Put 2cos⁡x=t⇒−2sin⁡xdx=dt\sqrt{2}\cos x = t \Rightarrow - \sqrt{2}\sin xdx = dt

I=−12∫dtt2−1=−12cos⁡h−1(2cos⁡x+c)I = \frac{- 1}{\sqrt{2}}\int_{}^{}\frac{dt}{\sqrt{t^{2} - 1}} = \frac{- 1}{\sqrt{2}}\cos h^{- 1}(\sqrt{2}\cos x + c).