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If \int_{0}^{\frac{\pi}{2}}\frac{d\theta}{9 \sin^{2}\theta + 4 \cos^{2}\theta}= k\pi,\, \, then\, \, the \, \, value \, \, of \, \, k\, \, is

  • Option 1)

    \frac{1}{16}

  • Option 2)

    \frac{1}{12}

  • Option 3)

    \frac{1}{8}

  • Option 4)

    \frac{1}{3}

 

Answers (1)

best_answer

As we discussed in concept

lower and upper limit -

\int_{a}^{b}f\left ( x \right )dx= \left ( F\left ( x \right ) \right )_{a}^{b}

                = F\left ( b \right )-F\left ( a \right )

 

- wherein

Where a is lower and b is upper limit.

 

 \int_{0}^{\frac{\pi }{2}}\frac{d\Theta }{gsin^2\Theta +4cos^2\Theta }=K\Pi

\int_{0}^{\frac{\pi }{2}}\frac{sec^2\Theta }{4+9 tan^2\Theta }d\Theta

\int_{0}^{\infty }\frac{dt}{4+9t^2}                              [\because tan \Theta =t , sec^2\Theta d\Theta =dt]

\frac{1}{6}tan^{-1}\frac{3t}{2}|^\infty_{0} = \frac{1}{6}\times \frac{\pi }{2} =\frac{\pi }{12}

\because k=\frac{1}{12}


Option 1)

\frac{1}{16}

Incorrect

Option 2)

\frac{1}{12}

Correct

Option 3)

\frac{1}{8}

Incorrect

Option 4)

\frac{1}{3}

Incorrect

Posted by

Aadil

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