400 Higher Engineering Mathematics
Problem 4. Evaluate∫ π
3
π
61
2cot^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
61
2cot^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(π3)2−π
3⎞
⎟
⎠−⎛
⎜
⎝−cot 2(π6)2−π
6⎞
⎟
⎠⎤
⎥
⎦=1
2[( 0. 2887 − 1. 0472 )−(− 0. 2887 − 0. 5236 )]= 0. 0269Now try the following exerciseExercise 155 Further problems on
integration ofsin^2 x,cos^2 x,tan^2 xandcot^2 xIn 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
3tan3θ−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
03sin^23 xdx[π
2or 1. 571]6.∫ π
4
0cos^24 xdx[π
8or 0. 3927]7.∫ 102tan^22 tdt [− 4 .185]8.∫ π
3
π
6cot^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+cWhenever 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 )+cand∫
sinnθcosθdθ =sinn+^1 θ
(n+ 1 )+cProblem 6. Evaluate∫ π
2
0sin^2 xcos^3 xdx.∫ π
2
0sin^2 xcos^3 xdx=∫ π
2
0sin^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
20=⎡
⎢
⎣(
sinπ
2) 33−(
sinπ
2) 55⎤
⎥
⎦−[0−0]=1
3−1
5=2
15or 0. 1333Problem 7. Evaluate∫ π
4
04cos^4 θdθ, correct to 4
significant figures.