Sunday, April 27, 2014

Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 38

Evaluate sin(lnx)dx by making a substitution first, then by using Integration by parts.
If we use z=lnx, then ez=x so dx=ezdz
Thus,


sin(lnx)dx=sinz(ezdz)=ezsinzdz
By using Integration by parts, if we let u=ez and dv=sinzdz then,
du=ezdz and v=cosz

Thus,

ezsinzdz=uvvdu=ezcosz(cosz)(ezdz)=ezcosz+ezcoszdz

Again by using Integration by parts, if we let

u1=ezanddv1=coszdz, thendu1=ezdzandv1=sinz

So,
ezcoszdz=u1v1v1du,=ezsinzsinz(ezdz)

Going back to the first equation,
ezsinzdz=ezcosz+[ezsinzsinz(ezdz)]

Combining like terms, we obtain

2ezsinzdz=ezcosz+ezsinzezsinzdz=ez(sinzcosz)2

but z=lnx,

Therefore,


ez(sinzcosz)2=elnx(sin(lnx)cos(lnx))2=x2[sin(lnx)cos(lnx)]+c

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