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7.2 Trigonometric Integrals
The three identities sin^2 x + cos^2 x = 1, cos^2 x = 12 (cos 2x + 1 ) and sin^2 x = 12 ( 1 − cos 2x) can be used to integrate expressions involving powers of Sine and Cosine. The basic idea is to use an identity to put your integral in a form where one of the substituions u = sin x or u = cos x may be applied.
Examples
- To compute
∫ sin^3 x dx, we first apply the identity sin^2 x + cos^2 x = 1 to get ∫ sin^3 x dx =
∫ sin x( 1 − cos^2 x) dx
This is now set up perfectly for the substitution u = cos x =⇒ du = − sin x dx: ∫ sin^3 x dx =
∫ ( 1 − u^2 )(− du) =
∫ u^2 − 1 du
= 1 3
u^3 − u + c = 1 3
cos^3 x − cos x + c
- The same trick works in the following: ∫ sin^3 x cos^2 x dx =
∫ sin x sin^2 x cos^2 x dx =
∫ sin x( 1 − cos^2 x) cos^2 x dx
=
∫ sin x(cos^2 x − cos^4 x) dx
=
∫ (u^2 − u^4 )(− du) (substitute u = cos x)
= − 13 u^3 + 15 u^5 + c = 15 cos^5 x − 13 cos^3 x + c
- With an even power, a double-angle formula is more useful. ∫ sin^2 x dx =
( 1 − cos 2x) dx = 1 2
x + 1 4
sin 2x + c
General Strategies
sinm^ x cosn^ x dx
- If n = 2 k + 1 is odd, replace cos^2 k^ x = ( 1 − sin^2 x)k, then ∫ sinm^ x cosn^ x dx =
∫ sinm^ x( 1 − sin^2 x)k^ cos x dx
Now substitute u = sin x =⇒ du = cos x dx ∫ sinm^ x cosn^ x dx =
∫ um( 1 − u^2 )k^ du
This is a polynomial in u, which can be integrated after multiplying out.
- If m is odd, repeat (1) with the roles of cos and sin reversed. Examples 1. and 2. above are of this form.
- If m = 2 k, n = 2 l are both even, write sin^2 x = 12 ( 1 − cos 2x) and cos^2 x = 12 ( 1 + cos 2x) to obtain ∫ sinm^ x cosn^ x dx = 1 2 k+l
∫ ( 1 − cos 2x)k( 1 + cos 2x)l^ dx
which yields integrals of powers of cos 2x. If these powers are odd, then we use the strategies above, if they are even we can repeat to find integrals in terms of cos 4x, etc. An alternative approach might utilise the identity sin x cos x = 12 sin 2x.
Examples
- To compute
sin^4 x dx we have to use two identities: sin^2 x = 12 ( 1 − cos 2x) and then cos^22 x = 1 2 (^1 +^ cos 4x). ∫ sin^4 x dx =
∫ (sin^2 x)^2 dx = 1 4
∫ ( 1 − cos 2x)^2 dx
= 1 4
∫ 1 − 2 cos 2x + cos^2 2 x dx
=
4 x^ −^
4 sin 2x^ +^
∫ cos^2 2 x dx (integrate the bits you can... )
=
4 x^ −^
4 sin 2x^ +^
∫ 1 + cos 4x dx (... use another identity)
=
8 x^ −^
4 sin 2x^ +^
32 sin 4x^ +^ c
- If both powers of sine and cosine are odd, you may use either of the first two strategies: ∫ sin^5 x cos^3 x dx =
∫ cos x sin^5 x( 1 − sin^2 x) dx =
∫ cos x(sin^5 x − sin^7 x) dx
=
∫ u^5 − u^7 du (substitute u = sin x)
= 1 6
sin^6 x − 1 8
sin^8 x + c 1
Alternatively ∫ sin^5 x cos^3 x dx =
∫ sin x( 1 − cos^2 x)^2 cos^3 x dx =
∫ sin x(cos^3 x − 2 cos^5 x + cos^7 x) dx
=
∫ u^3 − 2 u^5 + u^7 (− du) (let u = cos x)
= −
4 cos
(^4) x + 1 3 cos
(^6) x − 1 8 cos
(^8) x + c 2
Since both these answers are correct, we must have discovered a (nasty) trig identity! Indeed by evaluating both at x = 0 we can quickly see that c 1 = c 2 − 241 and the result is the evil-looking identity
4 sin^6 x − 3 sin^8 x = 1 − 6 cos^4 x + 8 cos^6 x − 3 cos^8 x
In cases other than the above we may be stuck: we can sometimes appeal to the anti-derivatives
∫ tan x dx = ln |sec x| + c,
∫ sec x dx = ln |sec x + tan x| + c
but these may be insufficient to solve the problem. Integration by parts or some other clever substi- tution may be necessary.
Examples
∫ tan^2 x sec^4 x dx =
∫ tan^2 x( 1 + tan^2 x) sec^2 x dx
=
∫ u^2 ( 1 + u^2 ) du (substitute u = tan x)
=
3 u
5 u
(^5) + c = 1 3 tan
(^3) x + 1 5 tan
(^5) x + c
∫ tan^5 x sec^3 x dx =
∫ tan^4 x sec^2 x sec x tan x dx =
∫ (sec^2 x − 1 )^2 sec^2 x sec x tan x dx
=
∫ (u^2 − 1 )^2 u^2 du (substitute u = sec x)
=
∫ u^6 − 2 u^4 + u^2 du = 1 7
u^7 − 2 5
u^5 + 1 3
u^3 + c
=
7 sec
(^7) x − 2 5 sec
(^5) x + 1 3 sec
(^3) x + c
∫ tan^2 x sec x dx =
∫ (sec^2 x − 1 ) sec x dx =
∫ sec^3 x − sec x dx
=
∫ sec^3 x dx − ln (^) |sec x + tan x|
∫ sec^3 x dx can be attacked by parts:
u = sec x dv = sec^2 x dx
du = sec x tan x dx v = tan x
∫ tan^2 x sec x dx = sec x tan x −
∫ sec x tan^2 x dx − ln |sec x + tan x|
=⇒
∫ tan^2 x sec x dx =
2 (sec^ x^ tan^ x^ −^ ln^ |sec^ x^ +^ tan^ x|) +^ c
Other Trigonometric Integrals
- Other trigonometric integrals will not tend to have general strategies: creativity is necessary!
- No other types of trigonometric integral are examinable in this class!
Suggested problems
- Evaluate the integrals:
(a)
∫ sin^3 x cos^2 x dx
(b)
∫ cos^4 ( 2 x) dx
- Evaluate the integrals:
(a)
∫ sec^4 x tan^3 x dx
(b)
∫ (^) π / 0
sec^3 x tan^3 x dx
- The region R 1 is bounded by the graph of y = tan x and the x-axis on the interval [0, π /3]. The region R 2 is bounded by the graph of y = sec x and the x-axis on the interval [0, π /6]. Which region has the greater area?
