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A comprehensive guide on how to integrate trigonometric functions using trigonometric identities and substitutions. It covers integrals of powers of trigonometric functions, integrals of products of sines and cosines, integrals of secants and tangents, and trigonometric substitutions. numerous examples to illustrate the concepts.
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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
sin^5 x 5
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
sin^3 x 3
sin^5 x 5
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
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
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
t 2
2 sin t cos t 4
t 2
sin t
1 − sin^2 t 2
sin−^1 x 2
x
1 − x^2 2
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
1 − sin^2 t + 13 (1 − sin^2 t)^3 /^2
9 − x^2 + 13 (9 − x^2 )^3 /^2 + C.
where x = 3 sin t, dx = 3 cos t dt.