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1.6. Trigonometric Integrals and Trigonometric Substitutions
1.6.1. Trigonometric Integrals. Here we discuss integrals of pow- ers of trigonometric functions. To that end the following half-angle identities will be useful:
sin^2 x =
(1 − cos 2x) ,
cos^2 x =
(1 + cos 2x).
Remember also the identities:
sin^2 x + cos^2 x = 1 , sec^2 x = 1 + tan^2 x.
1.6.1.1. Integrals of Products of Sines and Cosines. We will study now integrals of the form ∫ sinm^ x cosn^ x dx ,
including cases in which m = 0 or n = 0, i.e.: ∫ cosn^ x dx ;
sinm^ x dx.
The simplest case is when either n = 1 or m = 1, in which case the substitution u = sin x or u = cos x respectively will work.
Example:
sin^4 x cos x dx = · · ·
(u = sin x, du = cos x dx)
u^4 du =
u^5 5
+ C =
sin^5 x 5
+ C.
More generally if at least one exponent is odd then we can use the identity sin^2 x+cos^2 x = 1 to transform the integrand into an expression containing only one sine or one cosine.
Example: ∫ sin^2 x cos^3 x dx =
sin^2 x cos^2 x cos x dx
sin^2 x (1 − sin^2 x) cos x dx = · · ·
(u = sin x, du = cos x dx)
u^2 (1 − u^2 ) du =
(u^2 − u^4 ) du
u^3 3
u^5 5
+ C
sin^3 x 3
sin^5 x 5
+ C.
If all the exponents are even then we use the half-angle identities.
Example: ∫ sin^2 x cos^2 x dx =
1 2 (1^ −^ cos 2x)
1 2 (1 + cos 2x)^ dx
=
(1 − cos^2 2 x) dx
(1 − 12 (1 + cos 4x)) dx
(1 − cos 4x) dx
x 8
sin 4x 32
+ C.
1.6.1.2. Integrals of Secants and Tangents. The integral of tan x can be computed in the following way: ∫
tan x dx =
sin x cos x
dx = −
du u
= − ln |u| + C = − ln | cos x| + C ,
where u = cos x. Analogously ∫ cot x dx =
cos x sin x
dx =
du u
= ln |u| + C = ln | sin x| + C ,
where u = sin x.
(u = tan x, du = sec^2 x dx) ∫ tan^3 x ︸ ︷︷ ︸ u^3
sec^2 x dx ︸ ︷︷ ︸ du
u^3 du =
u^4 4
Although this answer looks different from the one obtained using the first method it is in fact equivalent to it because they differ in a con- stant:
1 4 tan
(^4) x = 1 4 (sec
(^2) x − 1) (^2) = 1 4 sec
(^4) x − 1 2 sec
(^2) x ︸ ︷︷ ︸ previous answer
+^14.
1.6.2. Trigonometric Substitutions. Here we study substitu- tions of the form x = some trigonometric function.
Example: Find
1 − x^2 dx.
Answer : We make x = sin t, dx = cos t dt, hence √ 1 − x^2 =
1 − sin^2 t =
cos^2 t = cos t ,
and ∫ (^) √ 1 − x^2 dx =
cos t cos t dt
cos^2 t dt
1 2 (1 + cos 2t)^ dt^ (half-angle identity)
=
t 2
sin 2t 4
+ C
t 2
2 sin t cos t 4
- C (double-angle identity)
t 2
sin t
1 − sin^2 t 2
+ C
sin−^1 x 2
x
1 − x^2 2
+ C.
The following substitutions are useful in integrals containing the following expressions:
expression substitution identity a^2 − u^2 u = a sin t 1 − sin^2 t = cos^2 t a^2 + u^2 u = a tan t 1 + tan^2 t = sec^2 t u^2 − a^2 u = a sec t sec^2 t − 1 = tan^2 t
So for instance, if an integral contains the expression a^2 −u^2 , we may try the substitution u = a sin t and use the identity 1 − sin^2 t = cos^2 t in order to transform the original expression in this way:
a^2 − u^2 = a^2 (1 − sin^2 t) = a^2 cos^2 t.
Example:
x^3 √ 9 − x^2
dx = 27
sin^3 t cos t √ 1 − sin^2 t
dt (x = 3 sin t)
sin^3 t dx
(1 − cos^2 t) sin t dx
− cos t +
cos^3 t 3
+ C
1 − sin^2 t + 13 (1 − sin^2 t)^3 /^2
+ C
9 − x^2 + 13 (9 − x^2 )^3 /^2 + C.
where x = 3 sin t, dx = 3 cos t dt.
