If-for-a-real-number-y-y-is-the-greatest-integer-less-than-or-equal-to-y-then-the-value-of-the-integeral-pi-2-3pi-2-2-sin-x-dx-is-

Question Number 27280 by kemhoney78@gmail.com last updated on 04/Jan/18

If for a real number y, [y] is the greatest

integer less than or equal to y, then the

value of the integeral \underset{π / 2}{\int^{3 π / 2}} [2 \sin x] d x is

Commented by abdo imad last updated on 04/Jan/18

l e t d o t h e c h a n g e m e n t x = \frac{π}{2} + t

\int_{\frac{π}{2}}^{\frac{3 π}{2}} [2 s i n x] d x = \int_{0}^{π} [2 c o s x] d x (2 c o s x = t \Leftrightarrow x = a r c o s (\frac{t}{2})) =

= \int_{2}^{- 2} [t] (- \frac{1}{2 \sqrt{1 - \frac{t^{2}}{4}}}) d t = \frac{1}{2} \int_{- 2}^{2} \frac{[t]}{\sqrt{1 - \frac{t^{2}}{4}}} d t

= \frac{1}{2} \int_{- 2}^{- 1} (\dots) d t + \frac{1}{2} \int_{- 1}^{0} (\dots) d t + \frac{1}{2} \int_{0}^{1} (\dots) d t + \frac{1}{2} \int_{1}^{2} (\dots) d t

= - \int_{- 2}^{- 1} \frac{d t}{\sqrt{1 - \frac{t^{2}}{4}}} - \frac{1}{2} \int_{- 1}^{0} \frac{d t}{\sqrt{1 - \frac{t^{2}}{4}}} + 0 + \frac{1}{2} \int_{1}^{2} \frac{d t}{\sqrt{1 - \frac{t^{2}}{4}}} d t

= - \int_{2}^{1} \frac{- d t}{\sqrt{1 - \frac{t^{2}}{4}}} + \frac{1}{2} \int_{1}^{2} \frac{d t}{\sqrt{1 - \frac{t^{2}}{4}}} - \frac{1}{2} \int_{1}^{0} \frac{- d t}{\sqrt{1 - \frac{t^{2}}{4}}}

= - \int_{1}^{2} \frac{d t}{\sqrt{1 - \frac{t^{2}}{4}}} + \frac{1}{2} \int_{1}^{2} \frac{d t}{\sqrt{1 - \frac{t^{2}}{2}}} - \frac{1}{2} \int_{0}^{1} \frac{d t}{\sqrt{1 - \frac{t^{2}}{4}}}

= - \frac{1}{2} \int_{1}^{2} \frac{d t}{\sqrt{1 - \frac{t^{2}}{2}}} - \frac{1}{2} \int_{0}^{1} \frac{d t}{\sqrt{1 - \frac{t^{2}}{4}}} d t

= - \frac{1}{2} \int_{0}^{2} \frac{d t}{\sqrt{1 - \frac{t^{2}}{4}}} d t = {[a r c o s (\frac{t}{2})]}_{0}^{2}

= a r c o s (1) - a r c o s 0 = - \frac{π}{2}

Answered by ajfour last updated on 04/Jan/18

I = \int_{π / 2}^{5 π / 6} d x + \int_{5 π / 6}^{π} (0) d x + \int_{π}^{7 π / 6} (- 1) d x

+ \int_{7 π / 6}^{3 π / 2} (- 2) d x

= \frac{π}{3} + 0 - \frac{π}{6} - \frac{2 π}{3} = - \frac{π}{2} .

Commented by ajfour last updated on 04/Jan/18

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