Question
The value of $$\int_{\frac{{{\pi ^3}}}{{27}}}^{\frac{{{\pi ^3}}}{8}} {\sin \,x\,dt,} $$ where $$t = {x^3},$$ is :
A.
$$\frac{{{\pi ^2}}}{6} + \left( {3 - \sqrt 3 } \right)\pi - 3$$
B.
$$\cos \frac{{{\pi ^3}}}{{27}} - \cos \frac{{{\pi ^3}}}{8}$$
C.
$$\frac{{{\pi ^2}}}{6}$$
D.
none of these
Answer :
$$\frac{{{\pi ^2}}}{6} + \left( {3 - \sqrt 3 } \right)\pi - 3$$
Solution :
$$\eqalign{
& I = \int_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\sin \,x.d\left( {{x^3}} \right)} = 3\int_{\frac{\pi }{3}}^{\frac{\pi }{2}} {{x^2}\sin \,x\,dx} \cr
& \int {{x^2}\sin \,x\,dx} = {x^2}\left( { - \cos \,x} \right) + \int {2x\cos \,x\,dx} \cr
& = - {x^2}\cos \,x + 2\left\{ {x\sin \,x - \int {\sin \,x\,dx} } \right\} \cr
& = - {x^2}\cos \,x + 2x\sin \,x + 2\cos \,x \cr
& \therefore I = 3\left[ { - {x^2}\cos \,x + 2x\sin \,x + 2\cos \,x} \right]_{\frac{\pi }{3}}^{\frac{\pi }{2}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = 3\pi - 3\left( { - \frac{{{\pi ^2}}}{9}.\frac{1}{2} + 2.\frac{\pi }{3}.\frac{{\sqrt 3 }}{2} + 2.\frac{1}{2}} \right) \cr
& = 3\pi + \frac{{{\pi ^2}}}{6} - \sqrt 3 \pi - 3 \cr
& = \frac{{{\pi ^2}}}{6} + \left( {3 - \sqrt 3 } \right)\pi - 3 \cr} $$