Understanding Engineering Mathematics

by the double angle formula for cos 2x.

=

1

2

∫

( 1 −cos 2x)dx

=

1

2

(

x−

sin 2x

2

)

+C

NB: Be careful with functions such as cos 2x, which, of course is not

the same thing as 2 cosx!

(ii) We can write

∫

cos^3 xdx=

∫

cos^2 xcosxdx

=

∫

(

1 −sin^2 x

)

cosxdx

Now putu=sinx, or equivalently:

=

∫

( 1 −sin^2 x)d(sinx)=sinx−

sin^3 x

3

+C

(iii) We can express sinAcosBas a sum of sines by using

sin(A+B)=sinAcosB+sinBcosA

sin(A−B)=sinAcosB−sinBcosA

Adding:

sinAcosB=^12 (sin(A+B)+sin(A−B))

So

sin 2xcos 3x=^12 (sin( 2 x+ 3 x)+sin( 2 x− 3 x))

=^12 (sin 5x+sin(−x))

=^12 (sin 5x−sinx)

So we can write:

∫

sin 2xcos 3xdx=

1

2

∫

(sin 5x−sinx)dx

=

1

2

(

−

cos 5x

5

+cosx

)

+C

=

1

2

cosx−

1

10

cos 5x+C

(iv) This is easy now, using the double angle formula for sin 2xback-

wards:

sinxcosx=^12 2sinxcosx

=^12 sin 2x