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  • I have a doubt, kindly clarify. - Integral Calculus - JEE Main-2

\int \frac{dx}{1+2\cos x}

  • Option 1)

    \frac{1}{\sqrt{3}}\ln \frac{\sqrt{3}+\tan _{\frac{x}{2}}}{\sqrt{3}-\tan _{\frac{x}{2}}}

  • Option 2)

    \frac{1}{2\sqrt{3}}\ln \frac{\sqrt{3}+\tan _{\frac{x}{2}}}{\sqrt{3}-\tan _{\frac{x}{2}}}

  • Option 3)

    \frac{2}{\sqrt{3}}\ln \frac{\sqrt{3}+\tan _{\frac{x}{2}}}{\sqrt{3}-\tan _{\frac{x}{2}}}

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we learned,

 

Type of Integration by perfect square -

 

The integrals are of the from 

(i)    \int \frac{1}{a\cos x+b\sin x}dx

(ii) \int \frac{1}{a+b\cos x}dx

(iii) \int \frac{1}{a+b\sin x}dx

 

- wherein

Working rule :

Resolve :

\cos x=\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}

and 

\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}

 

 

\int \frac{dx}{1+\frac{2-2\tan _{\frac{x}{2}}^{2}}{1+\tan _{\frac{x}{2}}^{2}}}=\int \frac{\sec _{\frac{x}{2}}^{2}dx}{3-\tan _{\frac{x}{2}}^{2}}

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


Option 1)

\frac{1}{\sqrt{3}}\ln \frac{\sqrt{3}+\tan _{\frac{x}{2}}}{\sqrt{3}-\tan _{\frac{x}{2}}}

Option 2)

\frac{1}{2\sqrt{3}}\ln \frac{\sqrt{3}+\tan _{\frac{x}{2}}}{\sqrt{3}-\tan _{\frac{x}{2}}}

Option 3)

\frac{2}{\sqrt{3}}\ln \frac{\sqrt{3}+\tan _{\frac{x}{2}}}{\sqrt{3}-\tan _{\frac{x}{2}}}

Option 4)

none of these

Posted by

gaurav

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