Friday, August 17, 2012

Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 42

Evaluate x2sin2xdx. Illustrate and check whether your answer is reasonable by gaphing both the function and its antiderivative suppose that c=0.

If we use 2=2x, then x=z2 so dx=12dz
x2sin2xdx=(z2)2(sinz)(dz2)=18z2sinzdz

By using integration by parts,
If we let u=z2 and dv=sinzdz. Then,
du=2zdz and v=cosz

Thus,


18z2sinzdz=uvvdu=z2cosz(cosz)(2zdz)=z2cosz+2zcoszdz

Again by using integration by parts, if we let u1=z and dv1=coszdz, then
du1=dz and v1=sinz


so, zcoszdz=u1v1vzdu1=zsinzsinzdz=zsinz(cosz)=zsinz+cosz


Going back to the first equation,

18z2sinzdz=18[z2cosz+2(zsinz+cosz)]=z2cosz8+zsinz4+cosz4+c


but z=2x, therefore,

x2sin2xdx=(2x)28cos(2x)+(2x)sin(2x)4+cos2x4+c=x2cos(2x)2+x2sin(2x)+cos2x4+c




We can see from the graph that our answer is reasonable, because the graph of the anti-derivative f is increasing when f is positive. On the other hand, the graph of f is decreasing when f is negative.

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