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The integral

\int_{0 }^{\pi }\sqrt{1+4\sin ^{2}\frac{x}{2}-4\sin \frac{x}{2}}dxequal to:

  • Option 1)

    4\sqrt{3}-4

  • Option 2)

    4\sqrt{3}-4-\frac{\pi }{3}

  • Option 3)

    \pi -4

  • Option 4)

    \frac{2\pi }{3}-4-4\sqrt{3}

 

Answers (1)

As learnt in concept

Fundamental Properties of Definite integration -

If the function is continuous in (a, b ) then integration of a function a to b will be same as the sum of integrals of the same function from a to c and c to b.

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

- wherein

 

 

 

= \int_{0 }^{\pi}\sqrt{\left (1-2sin\frac{x}{2} \right )^{2}}=\int_{0 }^{\pi}\left |1-2sin\frac{x}{2} \right |dx

=\int_{0 }^{\frac{\pi}{3}}\sqrt{\left (1-2sin\frac{x}{2} \right )}=\int_{\frac{\pi}{3}}^{\pi}\left (1-2sin\frac{x}{2} \right )dx 

=\left (x+4cos\frac{x}{2} \right )^{\frac{\pi}{3}}_{0}- \left (x+4cos\frac{x}{2} \right )^{\pi}_{\frac{\pi}{3}}

=4\sqrt{3}-4-\frac{\pi}{3}


Option 1)

4\sqrt{3}-4

This is incorrect

Option 2)

4\sqrt{3}-4-\frac{\pi }{3}

This is correct

Option 3)

\pi -4

This is incorrect

Option 4)

\frac{2\pi }{3}-4-4\sqrt{3}

This is incorrect

Posted by

Sabhrant Ambastha

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