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Question

Question: \[\int_{}^{}\frac{dx}{3 + 4\cos x} =\]...

∫dx3+4cos⁡x=\int_{}^{}\frac{dx}{3 + 4\cos x} =

A

17log⁡(tan⁡(x/2)−7tan⁡(x/2)+7)+c\frac{1}{\sqrt{7}}\log\left( \frac{\tan(x/2) - \sqrt{7}}{\tan(x/2) + \sqrt{7}} \right) + c

B

17log⁡(tan⁡(x/2)+7tan⁡(x/2)−7)+c\frac{1}{\sqrt{7}}\log\left( \frac{\tan(x/2) + \sqrt{7}}{\tan(x/2) - \sqrt{7}} \right) + c

C

17log⁡(7+tan⁡(x/2)7−tan⁡(x/2))+c\frac{1}{\sqrt{7}}\log\left( \frac{\sqrt{7} + \tan(x/2)}{\sqrt{7} - \tan(x/2)} \right) + c

D

17log⁡(7−tan⁡(x/2)7+tan⁡(x/2))+c\frac{1}{\sqrt{7}}\log\left( \frac{\sqrt{7} - \tan(x/2)}{\sqrt{7} + \tan(x/2)} \right) + c

Answer

17log⁡(tan⁡(x/2)+7tan⁡(x/2)−7)+c\frac{1}{\sqrt{7}}\log\left( \frac{\tan(x/2) + \sqrt{7}}{\tan(x/2) - \sqrt{7}} \right) + c

Explanation

Solution

If b>ab > a then,

∫dxa+bcos⁡x=1b2−a2log⁡[b−atan⁡x2+b+ab−atan⁡x2−b+a]+c\int_{}^{}\frac{dx}{a + b\cos x} = \frac{1}{\sqrt{b^{2} - a^{2}}}\log\left\lbrack \frac{\sqrt{b - a}\tan\frac{x}{2} + \sqrt{b + a}}{\sqrt{b - a}\tan\frac{x}{2} - \sqrt{b + a}} \right\rbrack + c ⇒I=17log⁡(tan⁡x2+7tan⁡x2−7)+c\Rightarrow I = \frac{1}{\sqrt{7}}\log\left( \frac{\tan\frac{x}{2} + \sqrt{7}}{\tan\frac{x}{2} - \sqrt{7}} \right) + c