400 Higher Engineering Mathematics
Problem 4. Evaluate
π
3
π
6
1
2
cot
2 2θ dθ.
Since cot
2
θ +1 = cosec
2
θ, then cot
2
θ = cosec
2
θ−1
and cot 2 2θ = cosec 2 2θ − 1.
Hence
π
3
π
6
1
2
cot
2 2θ dθ
=
1
2
π
3
π
6
(cosec
2 2θ − 1) dθ =
1
2
−cot 2θ
2
− θ
π
3
π
6
=
1
2
⎡
⎢
⎣
⎛
⎜
⎝
−cot 2
π
3
2
−
π
3
⎞
⎟
⎠−
⎛
⎜
⎝
−cot 2
π
6
2
−
π
6
⎞
⎟
⎠
⎤
⎥
⎦
=
1
2
[(0.2887 − 1.0472) − (−0.2887 − 0.5236)]
= 0.0269
Now try the following exercise
Exercise 155 Further problems on
integration of sin
2 x, cos 2 x, tan 2 x and cot 2 x
In Problems 1 to 4, integrate with respect to the
variable.
1. sin
2 2x
1
2
x −
sin 4x
4
+ c
2. 3 cos 2 t
3
2
t +
sin 2t
2
+ c
3. 5 tan 2 3θ
5
1
3
tan 3θ − θ
+ c
4. 2 cot 2 2t
[−(cot 2t + 2t ) + c]
In Problems 5 to 8, evaluate the definite integrals,
correct to 4 significant figures.
5.
π
3
0
3 sin
2 3x dx
π
2
or 1.571
6.
π
4
0
cos
2 4x dx
π
8
or 0.3927
7.
1
0
2 tan
2 2t dt
[−4.185]
8.
π
3
π
6
cot
2
θ dθ
[0.6311]
40.3 Worked problems on powers of
sines and cosines
Problem 5. Determine
sin 5 θ dθ.
Since cos 2 θ + sin
2
θ = 1 then sin
2
θ = (1 − cos 2 θ).
Hence
sin
5
θ dθ
=
sin θ(sin
2
θ)
2 dθ =
sin θ(1 − cos
2
θ)
2 dθ
=
sin θ(1 − 2 cos
2
θ + cos
4
θ)dθ
=
(sin θ − 2 sin θ cos
2
θ + sin θ cos
4
θ)dθ
= −cos θ +
2 cos 3 θ
3
−
cos 5 θ
5
+ c
Whenever a power of a cosine is multiplied by a sine of
power 1, or vice-versa, the integral may be determined
by inspection as shown.
In general,
cos
n
θ sin θ dθ =
−cos n+1 θ
(n + 1)
+ c
and
sin
n
θ cos θ dθ
=
sin n+1 θ
(n + 1)
+ c
Problem 6. Evaluate
π
2
0
sin
2 x cos
3 x dx.
π
2
0
sin
2 x cos
3 x dx =
π
2
0
sin
2 x cos
2 x cos x dx
=
π
2
0
(sin
2 x)(1 − sin
2 x)(cos x) dx
=
π
2
0
(sin
2 x cos x − sin
4 x cos x) dx
=
sin 3 x
3
−
sin 5 x
5
π
2
0
=
⎡
⎢
⎣
sin
π
2
3
3
−
sin
π
2
5
5
⎤
⎥
⎦ − [0 − 0]
=
1
3
−
1
5
=
2
15
or 0.1333
Problem 7. Evaluate
π
4
0
4 cos
4
θ dθ, correct to 4
significant figures.