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Question: \[\int_{}^{}{\frac{\mathbf{\cos}\mathbf{x}\mathbf{dx}}{\mathbf{\cos}\mathbf{3}\mathbf{x}}\mathbf{=}}...

∫cos⁡xdxcos⁡3x=\int_{}^{}{\frac{\mathbf{\cos}\mathbf{x}\mathbf{dx}}{\mathbf{\cos}\mathbf{3}\mathbf{x}}\mathbf{=}}

A

123log⁡1+3tan⁡x1−3tan⁡x+c\frac{1}{2\sqrt{3}}\log\frac{1 + \sqrt{3}\tan x}{1 - \sqrt{3}\tan x} + c

B

−123log⁡1+3tan⁡x1−3tan⁡x+c\frac{- 1}{2\sqrt{3}}\log\frac{1 + \sqrt{3}\tan x}{1 - \sqrt{3}\tan x} + c

C

16tan⁡−1(3tan⁡x)+c\frac{1}{\sqrt{6}}\tan^{- 1}(\sqrt{3}\tan x) + c

D

None of these

Answer

123log⁡1+3tan⁡x1−3tan⁡x+c\frac{1}{2\sqrt{3}}\log\frac{1 + \sqrt{3}\tan x}{1 - \sqrt{3}\tan x} + c

Explanation

Solution

Let I=∫cos⁡xcos⁡3xdx=∫cos⁡x4cos⁡3x−3cos⁡xdxI = \int_{}^{}\frac{\cos x}{\cos 3x}dx = \int_{}^{}\frac{\cos x}{4\cos^{3}x - 3\cos x}dxDividing the numerator and denominator by cos⁡2x,\cos^{2}x, we have I =∫dx4cos⁡2x−3= \int_{}^{}\frac{dx}{4\cos^{2}x - 3}

I=∫dt1−3t2=13∫dt13−t2=13∫dt(13)2−t2=13.12.13log⁡∣13+t13−t∣+c=123log⁡∣1+3t1−3t∣+cI = \int_{}^{}{\frac{dt}{1 - 3t^{2}} = \frac{1}{3}\int_{}^{}{\frac{dt}{\frac{1}{3} - t^{2}} = \frac{1}{3}\int_{}^{}\frac{dt}{\left( \frac{1}{\sqrt{3}} \right)^{2} - t^{2}}}} = \frac{1}{3}.\frac{1}{2.\frac{1}{\sqrt{3}}}\log\left| \frac{\frac{1}{\sqrt{3}} + t}{\frac{1}{\sqrt{3}} - t} \right| + c = \frac{1}{2\sqrt{3}}\log\left| \frac{1 + \sqrt{3}t}{1 - \sqrt{3}t} \right| + c

I =123log⁡∣1+3tan⁡x1−3tan⁡x∣+c.= \frac{1}{2\sqrt{3}}\log\left| \frac{1 + \sqrt{3}\tan x}{1 - \sqrt{3}\tan x} \right| + c.