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  • Solve this problem - The value of, where [t] denotes the greatest integer less than equal to t, is: - Integral Calculus - JEE Main

The value of \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{[x] + [\sin x] + 4}, where [t] denotes the greatest integer less than equal to t, is:

  • Option 1)

    \frac{3}{10}(4\pi-3)

  • Option 2)

    \frac{3}{20}(4\pi-3)

  • Option 3)

    \frac{1}{12}(7\pi - 5)

  • Option 4)

    \frac{1}{12}(7\pi + 5)

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_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{[x]+[\sin x]+4}

from the concept we have learnt 

I=\int_{\frac{-\pi}{2}}^{-1}\frac{dx}{-2-1+4}+\int_{-1}^{0}\frac{dx}{-1-1+4}+\int_{0}^{1}\frac{dx}{0+0+4}+\int_{1}^{\frac{\pi}{2}}\frac{dx}{1+0+4}

I=\int_{\frac{-\pi}{2}}^{-1}\frac{dx}{1}+\int_{-1}^{0}\frac{dx}{2}+\int_{0}^{1}\frac{dx}{4}+\int_{1}^{\frac{\pi}{2}}\frac{dx}{5}

I=(-1+\frac{\pi}{2})+\frac{1}{2}(0+1)+\frac{1}{4}(1)+\frac{1}{5}(\frac{\pi}{2}-1)

    =\frac{-9}{20}+\frac{3\pi}{5}

 

 


Option 1)

\frac{3}{10}(4\pi-3)

Option 2)

\frac{3}{20}(4\pi-3)

Option 3)

\frac{1}{12}(7\pi - 5)

Option 4)

\frac{1}{12}(7\pi + 5)

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