Evaluate `int_(-1)^(3)(x-[x])dx` ,where [.] denotes the greatest integer function.

Answer

4

Explanation

`I=int_(0)^(oo)(sin5x)/x dx`
Let `5x=t`
`:. dx=1/5dt`
`:. I=int_(0)^(oo) (sin(t))/(t//5) . (dt)/5=int_(0)^(oo) (sint)/t dt=alpha`
b. `I=int_(0)^(sin^(2)x)dx`
`=int_(0)^(oo) (sin^(2)x)(1/(x^(2)))dx`
`=[sin^(2)x((-1)/x)]_(0)^(oo) -int_(0)^(oo) (2sinxcosx)xx((-1)/x)dx`
`=[(-sin^(2)x)/x]_(0)^(oo) +int_(0)^(oo) (sin2x)/x dx` put `2x=t`
`=lim_(xto oo) [-(sin^(2)x)/x]-lim_(xto oo) [-(sin^(2)x)/x]+alpha`
`=0-0+alpha=alpha`
c. `I=int_(0)^(oo) (sin^(3)x)/x dx`
`I=1/4[3int_(0)^(oo) (sinx)/x dx-int_(0)^(oo) (sin3x)/x dx]`
`=1/4[3alpha-alpha]=(alpha)/2`
d. `int_(0)^(oo) (sink_(1)xcosk_(2)x)/xdx`
`=1/2int_(0)^(oo) (2sink_(1)xcosk_(2)x)/xdx`
`=1/2int_(0)^(oo) (sin(k_(1)+k_(2))x+sin(k_(1)-k_(2))x)/x dx`
`=1/2 (alpha+alpha)=alpha`
`:. int_(0)^(oo) (sin(k_(1)x).cosk_(2)x)/x dx-alpha=0`