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  • How to solve this problem- - Integral Calculus - JEE Main-6

\int \frac{dx}{3+4\cos ^{2}x}

  • Option 1)

    \frac{\sqrt{3}}{\sqrt{7}}\tan ^{-1}\sqrt{\frac{3}{7}}\tan x+C

  • Option 2)

    \frac{1}{\sqrt{21}}\tan ^{-1}\sqrt{\frac{3}{7}}\tan x+C

  • Option 3)

    \sqrt{3}\tan ^{-1}\sqrt{\frac{3}{7}}\tan x+C

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we learned,

Type of Integration by perfect square -

Integration are in the format 

(i) \int \frac{dx}{a^{2}\sin ^{2}x+2b\sin x\cos x+c^{2}\cos^{2}x}

(ii) \int \frac{dx}{a\cos^{2}x+b} (iii) \int \frac{dx}{a+b\sin^{2}x}

- wherein

Working rule :

Divide by cos^{2}x in each case and pull t=\tan x ,  dt=\sec^{2}xdx

 

 

 Divide by cos^{2}x

 

\int \frac{\sec ^{2}xdx}{3\sec ^{2}x+4}=\int \frac{\sec ^{2}xdx}{7+3\tan ^{2}x}

Put \tan x=t\: \Rightarrow \: \sec ^{2}xdx=dt

\int \frac{dt}{7+3t^{2}}=\frac{1}{3}\int \frac{dt}{t^{2}+\left ( \sqrt{\left (\frac{7}{3} \right )} \right )^{2}}

=\frac{1}{3}\times \frac{\sqrt{3}}{\sqrt{7}}\tan ^{-1}\sqrt{\frac{3}{7}}\tan x+C


Option 1)

\frac{\sqrt{3}}{\sqrt{7}}\tan ^{-1}\sqrt{\frac{3}{7}}\tan x+C

Option 2)

\frac{1}{\sqrt{21}}\tan ^{-1}\sqrt{\frac{3}{7}}\tan x+C

Option 3)

\sqrt{3}\tan ^{-1}\sqrt{\frac{3}{7}}\tan x+C

Option 4)

none of these

Posted by

gaurav

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