   Chapter 8.3, Problem 14E

Chapter
Section
Textbook Problem

Finding an Indefinite Integral Involving Sine and Cosine In Exercises 3–14, find the indefinite integral. ∫ x 2 sin 2 x d x

To determine

To calculate: the value of the indefinite integral x2sin2xdx

Explanation

Given: The indefinite integral x2sin2xdx

Formula Used:

cos2t+sin2t=1.

cos2x=12sin2x.

cos2xdx=12sin2x+C.

(sin2x)dx=12cos2x+C.

ddxxn=nxn1.

xndx=xn+1n+1+c

Calculation:

Integrating by parts and letting u=x2 ----------(1), and

dv=sin2xdx ---------(2)

Applying the trigonometric identity cos2t=12sin2t, equation (2) becomes

dv=(1cos2x2)dx ------------(3)

Integrating equation (3), we get

dv=(1cos2x2)dxdv=12dx12(cos2x)dx

dv=x214sin2x+C ---------------(4)

Now,

x2sin2xdx=x2sin2xdx(ddxx2sin2xdx)dx.

x2sin2xdx=x2dv(2xdv)dx -------(5)

Putting the result of (4) in equation (5), we get

x2sin2xdx=x2(x214sin2x+C)(2x(x214sin2x+C))dx.

x2sin2xdx=x32x24sin2x(x2x2sin2x)dx

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