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  • How to solve this problem- - Integral Calculus - JEE Main-2

What is the value of  \int_{\Pi /6}^{\Pi /3}\frac{dx}{1+\sqrt{tanx}} ?

  • Option 1)

    \Pi /6

  • Option 2)

    \Pi /4

  • Option 3)

    \Pi /12

  • Option 4)

    \Pi /3

 

Answers (1)

As we learnt

Properties of Definite integration -

\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx

When \int_{0}^{b}f\left ( x \right )dx= \int_{0}^{b}f\left ( b-x \right )dx

 

- wherein

Put the \left ( a+b-x \right ) at the place of x in f\left ( x \right )

 

 

Let I = \int_{\pi /6}^{\pi/3}\frac{dx}{1+\sqrt{tanx}} 

                        = \int_{\pi /6}^{\pi/3}\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx

                        = \int_{\pi /6}^{\pi/3}\frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx   (since  a + b = p/2)

                        Hence,2I=\int_{\pi /6}^{\pi/3}dx=\pi/3-\pi/6=\pi/6 

                        Þ  I=\pi/12


Option 1)

\Pi /6

Option 2)

\Pi /4

Option 3)

\Pi /12

Option 4)

\Pi /3

Posted by

Vakul

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