We are at Part 13 of a 21 part series. This time we will work with integrals involving trigonometric identities.
Problems
All problems today will be indefinite integrals. Problems with definite integrals will be handled similarly.
Identities Involving
sin^2 x + cos^2 x = 1
cos (2x) = 1 - 2 sin^2 x
sin (2x) = 2 sin x cos x
sin^2 x = (1 - cos 2x)/2
cos^2 x = (1 + cos 2x)/2
1. ∫ cos^2 x dx
∫ cos^2 x dx
= ∫ 1 - sin^2 x dx
= ∫ 1 dx - ∫ sin^2 x dx
= x - ∫ 1/2 - 1/2 * cos 2x dx
= x - ( ∫ 1/2 dx - ∫ 1/2 * cos 2x dx )
= x - 1/2* x + 1/2 * ∫ cos 2x dx
= 1/2 * x + 1/2 * sin 2x * 1/2
= 1/2 * x + 1/4 * sin 2x
Final: ∫ cos^2 x dx = 1/2 * x + 1/4 * sin 2x + C
Note:
∫ cos (ax) dx = sin (ax)/a
∫ sin (ax) dx = -cos (ax)/a
2. ∫ sin^2 x cos^2 x dx
∫ sin^2 x cos^2 x dx
= ∫ (1 - cos 2x)/2 * (1 + cos 2x)/2 dx
= ∫ 1/4 * (1^2 - cos^2 (2x)) dx
= 1/4 * ( ∫ 1 - cos^2 (2x) dx )
= 1/4 * ( ∫ 1 - (1 + cos 4x)/2) dx )
= 1/4 * ( ∫ 1/2 - 1/2 * cos 4x dx)
= 1/4 * (1/2 * x - 1/8 * sin 4x)
Final: ∫ sin^2 x cos^2 x dx = 1/4 * (1/2 *x - 1/8 * sin 4x) + C
Trigonometric Substitutions
If the integrand has...
√(a^2 - x^2): use the substitution x = a sin u, dx = a cos u du
√(a^2 + x^2): use the substitution x = a tan u, dx = a sec^2 u du
√(x^2 - a^2): use the substitution x = a sec u, dx = a tan u sec u du
These can get messy!
Identity: 1 + tan^2 x = sec^2 x
3. ∫ 1 / √(4 - x^2) dx
a^2 = 4
a = 2
x = 2 sin u
dx = 2 cos u du
∫ 1/√(4 - x^2) dx
= ∫ (2 cos u) / √(4 - 4 sin^2 u) du
= ∫ (2 cos u) / (√4 * √(1 - sin^2 u) ) du
= ∫ (2 cos u) / (2 * √( cos^2 u ) ) du
= ∫ (2 cos u) / (2 * cos u) du
= ∫ 1 du
= u
x = 2 sin u implies that u = asin(x/2)
= asin(x/2)
Final: ∫ 1 / √(4 - x^2) dx = asin(x/2) + C
4. ∫ 1/ √(x^2 + 64) dx
a^2 = 64
a = 8
x = 8 tan u
dx = 8 sec^2 u du
u = atan(x/8)
∫ 1/ √(x^2 + 64) dx
= ∫ (8 sec^2 u)/(√(64 * tan^2 u + 64) du
= ∫ (8 sec^2 u)/(8 * √(sec^2 u)) du
= ∫ (8 sec^2 u)/(8 * sec u) du
= ∫ sec u du
∫ sec x dx = ln |sec x + tan x|
= ln |sec u + tan u|
= ln | sec(atan(x/8)) + tan(atan(x/8)) |
Refer to the right triangle below:
u = atan (x/8)
tan u = x/8
cos u = 8/√(x^2 + 64)
sec u = √(x^2 + 64)/8
Then:
ln | sec(atan(x/8)) + tan(atan(x/8)) |
= ln | √(x^2 + 64)/8 + x/8 |
= ln | √(x^2 + 64) + x | - ln 8
Final:
∫ 1 / √(x^2 + 64) dx
= ln | √(x^2 + 64) + x | - ln 8 + C
Note that ln 8 + C is a constant. Then.
∫ 1 / √(x^2 + 64) dx
= ln | √(x^2 + 64) + x | + C
5. ∫ 1 / (x^2 - 25)^(3/2) dx
a^2 = 25
a = 5
x = 5 sec u
u = asec(x/5) = acos(5/x)
dx = 5 sec u tan u du
1 + tan^2 x = sec^ x
Then:
∫ 1 / (x^2 - 25)^(3/2) dx
= ∫ (5 sec u tan u) / ( 25 sec^2 u - 25 )^(3/2) du
= ∫ (5 sec u tan u) / ( 25 tan^2 u )^(3/2) du
= ∫ (5 sec u tan u) / ( 25^(3/2) * tan^3 u ) du
= 5 / 25^(3/2) * ∫ (sec u tan u) / tan^3 u du
= 1/25 * ∫ cos u/sin^2 u du
= 1/25 * ∫ cot u csc u du
∫ cot x csc x dx = -csc x
5 / 25^(3/2) = 5 / ((5^2)^(3/2)) = 5 / (5^3) = 1 / 5^2 = 1/25
= 1/25 * -csc u
= -1/25 * csc(acos(5/x))
Refer to the triangle below:
x = 5 sec u
x/5 = sec u
5/x = cos u
√(x^2 - 25)/x = sin u
x/√(x^2 - 25) = csc u
Then:
-1/25 * csc(acos(5/x))
= -1/25 * x/√(x^2 - 25)
Final:
∫ 1 / (x^2 - 25)^(3/2) dx = -1/25 * x / √(x^2 - 25) + C
That concludes Part 13 of our series. Next time we will work with decomposition of fractions.
Have a good day,
Eddie
This blog is property of Edward Shore. © 2012
