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Mathematics Question on Definite Integral

The value of ∫−π/2π/2dx[x]+[sin⁡x]+4\int\limits^{\pi/2}_{-\pi/2} \frac{dx}{\left[x\right]+\left[\sin x\right]+4} where [t] denotes the greatest integer less than or equal to t, is :

A

112(7π+5)\frac{1}{12} (7 \pi + 5)

B

310(4π−3)\frac{3}{10} (4\pi - 3)

C

112(7π−5)\frac{1}{12} (7\pi - 5)

D

320(4π−3)\frac{3}{20} (4\pi - 3)

Answer

320(4π−3)\frac{3}{20} (4\pi - 3)

Explanation

Solution

I=∫−π2π2dx[x]+[sin⁡x]+4I = \int^{\frac{\pi}{2}}_{\frac{-\pi}{2}} \frac{dx}{\left[x\right]+\left[\sin x\right]+4}
=∫−π2−1dx−2−1+4+∫−10dx−1−1+4+∫01dx0+0+4= \int^{-1}_{\frac{-\pi}{2}} \frac{dx}{-2-1+4} + \int^{0}_{-1} \frac{dx}{-1-1+4} +\int^{1}_{0} \frac{dx}{0+0+4}
+∫1π2dx1+0+4+ \int^{\frac{\pi}{2}}_{1} \frac{dx}{1+0+4}
∫−π2−1dx1+∫−10dx2+∫01dx4+∫1π2dx5\int^{-1}_{\frac{-\pi}{2}} \frac{dx}{1} +\int^{0}_{-1} \frac{dx}{2} +\int^{1}_{0} \frac{dx}{4} + \int^{\frac{\pi}{2}}_{1} \frac{dx}{5}
(1−+π2)+12(0+1)+14+15(π2−1)\left(1- + \frac{\pi}{2}\right) + \frac{1}{2} \left(0+1\right) + \frac{1}{4} + \frac{1}{5} \left(\frac{\pi}{2} -1\right)
−1+12+14−15+π2+π10-1 + \frac{1}{2} +\frac{1}{4}- \frac{1}{5}+ \frac{\pi}{2}+\frac{\pi}{10}
−20+10+5−420+6π10\frac{-20+10+5-4}{20} + \frac{6\pi}{10}
−920+3π5\frac{-9}{20} +\frac{3\pi}{5}