Page 225 - Calculus Demystified

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EXAMPLE 7.13 CHAPTER 7 Methods of Integration
Calculate
π/2

4
4
sin x cos xdx.
0
SOLUTION
Substituting
1 − cos 2x 1 + cos 2x
2 2
sin x = and cos x =
2 2
into the integrand yields
π/2
2 2
1 − cos 2x 1 + cos 2x
· dx
2 2
0
π/2
1
2
4
= 1 − 2 cos 2x + cos 2xdx.
16 0
Again using formula II, we find that our integral becomes
1 π/2 1 + cos 4x 2
1 −[1 + cos 4x]+ dx
16 0 2
π/2
1 1
2
= 1 −[1 + cos 4x]+ 1 + 2 cos 4x + cos 4x dx.
16 0 4
Applying formula II one last time yields
π/2
1 1 1 + cos 8x
1 −[1 + cos 4x]+ 1 + 2 cos 4x + dx
16 0 4 2
π/2
1 1 1 1 x sin 8x
= − sin 4x + x + sin 4x + +
16 4 4 2 2 16
0

1 1 π π 1

= −0 + + 0 + + 0 − −0 + (0 + 0 + 0 + 0)
16 4 2 4 4
3π
= .
256
You Try It: Calculate the integral
π/3

3
3
sin s cos sds.
π/4
You Try It: Calculate the integral
π/3

2
4
sin s cos sds.
π/4