The definite integral of x times SiNx The interval is [- 1,1]

The definite integral of x times SiNx The interval is [- 1,1]

Integration by parts
∫xsinxdx=-xcosx+∫cosxdx
=-Xcosx + SiNx + C (C is the integral constant)

Find the definite integral ∫ 1 − 1F (x) DX, where f (x) = SiNx − 1 & nbsp; & nbsp; (x ≤ 0) x2 & nbsp; & nbsp; (x > 0)

∫1−1f（x）dx=∫0−1（sinx-1）dx+∫10x2dx=（-cosx-x）|0−1+13x3|10=cos1-2+13=cos1-53

Finding the definite integral f π 0 xsinx SiNx where x is square

∫π 0(xsinx2dx)
=∫π 0(sinx2dx2*1/2)
=1/2∫π 0(sinx2dx2)
=1/2[-cosx2]( π 0)
=1/2[-cosπ2-(-cos02)]
=1/2(cos02-cosπ2)
=1/2-1/2cosπ2
Except 1 / 2 for half, all the other 2 are squares

Y = (1-cosx) / xsinx

xsin(x)sin(x)-(1-cos(x))cox(x)/(xsin(x))∧2