400 Higher Engineering Mathematics
Problem 4. Evaluate
∫ π
3
π
6
1
2
cot^22 θdθ.
Since cot^2 θ+ 1 =cosec^2 θ, then cot^2 θ=cosec^2 θ− 1
and cot^22 θ=cosec^22 θ−1.
Hence
∫ π
3
π
6
1
2
cot^22 θdθ
=
1
2
∫ π
3
π
6
(cosec^22 θ− 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 ofsin^2 x,cos^2 x,tan^2 xandcot^2 x
In Problems 1 to 4, integrate with respect to the
variable.
- sin^22 x
[
1
2
(
x−
sin4x
4
)
+c
]
- 3cos^2 t
[
3
2
(
t+
sin2t
2
)
+c
]
- 5tan^23 θ
[
5
(
1
3
tan3θ−θ
)
+c
]
- 2cot^22 t [−(cot 2t+ 2 t)+c]
In Problems 5 to 8, evaluate the definite integrals,
correct to 4 significant figures.
5.
∫ π
3
0
3sin^23 xdx
[π
2
or 1. 571
]
6.
∫ π
4
0
cos^24 xdx
[π
8
or 0. 3927
]
7.
∫ 1
0
2tan^22 tdt [− 4 .185]
8.
∫ π
3
π
6
cot^2 θdθ [0.6311]
40.3 Worked problems on powers of
sines and cosines
Problem 5. Determine
∫
sin^5 θdθ.
Since cos^2 θ+sin^2 θ=1thensin^2 θ=( 1 −cos^2 θ).
Hence
∫
sin^5 θdθ
=
∫
sinθ(sin^2 θ)^2 dθ=
∫
sinθ( 1 −cos^2 θ)^2 dθ
=
∫
sinθ( 1 −2cos^2 θ+cos^4 θ)dθ
=
∫
(sinθ−2sinθcos^2 θ+sinθcos^4 θ)dθ
=−cosθ+
2cos^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,
∫
cosnθsinθdθ=
−cosn+^1 θ
(n+ 1 )
+c
and
∫
sinnθcosθdθ =
sinn+^1 θ
(n+ 1 )
+c
Problem 6. Evaluate
∫ π
2
0
sin^2 xcos^3 xdx.
∫ π
2
0
sin^2 xcos^3 xdx=
∫ π
2
0
sin^2 xcos^2 xcosxdx
=
∫ π
2
0
(sin^2 x)( 1 −sin^2 x)(cosx)dx
=
∫ π
2
0
(sin^2 xcosx−sin^4 xcosx)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
4cos^4 θdθ, correct to 4
significant figures.