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  • Solve it, - The value of the integral (where denotes the greatest integer less than or equal to) is : - Integral Calculus - JEE Main

The value of the integral \int_{-2}^{2}\frac{\sin ^{2}x}{\begin{bmatrix} \frac{x}{\pi} \end{bmatrix}+\frac{1}{2}}dx

(where \begin{bmatrix} x \end{bmatrix}  denotes the greatest integer less than or equal to x )  is :

  • Option 1)

    \sin 4

  • Option 2)

    0

  • Option 3)

    4

  • Option 4)

    4-\sin 4

Answers (1)

best_answer

 

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

 

 

I=\int_{2}^{-2}\frac{\sin ^{2}x}{[\frac{x}{\pi }]+\frac{1}{2}}dx

=\int_{2}^{0}\frac{\sin ^{2}x}{\frac{1}{2}}dx+\int_{0}^{-2}\frac{\sin ^{2}x}{-1+\frac{1}{2}}dx

2\oint_{2}^{0}\sin ^{2}xdx 1-2\int_{0}^{-2}\sin ^{2}xdx

Putx=-p\Rightarrow dx=-dp\sin ^{2}(-p)=\sin ^{2}p

=2\oint_{0}^{2}\sin ^{2}xdx+2\int_{2}^{0}\sin ^{2}Pdp

=0

 

 

 


Option 1)

\sin 4

Option 2)

0

Option 3)

4

Option 4)

4-\sin 4

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