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A lecture on trigonometric integrals and trigonometric substitution∗ alozano† 2013-03-21 19:58:03 1 Trigonometric Integrals First, we must recall a few trigonometric identities: sin2 x + cos2 x 2 sec x = 1 (1) 2 1 + tan x 1 − cos(2x) sin2 x = 2 1 + cos(2x) cos2 x = 2 sin(2x) = 2 sin x cos x cos(2x) = = 2 2 cos x − sin x. (2) (3) (4) (5) (6) The most usual integrals which involve trigonometric functions can be solved using the identities above. R R Example 1.1. sin xdx = − cos x+C and cos xdx = sin x+C are immediate integrals. R R Example 1.2. For sin2 xdx, cos2 xdx we use formulas (3) and (4) respectively, e.g. Z Z Z 1 − cos(2x) 1 1 sin(2x) sin2 xdx = dx = (1 − cos(2x))dx = x− + C. 2 2 2 2 R R Example 1.3. For integrals of the form cosm x sin x dx or sinm x cos x dx we use substitution with u = cos x or u = sin x respectively, e.g. Z Z u3 cos3 x cos2 x sin xdx = −u2 du = − +C = − +C. [u = cos x, du = − sin xdx] 3 3 ∗ hALectureOnTrigonometricIntegralsAndTrigonometricSubstitutioni created: h2013-0321i by: halozanoi version: h37576i Privacy setting: h1i hFeaturei h26A36i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 In the following examples, we use equations (1) in the forms sin2 x = 1 − cos x or cos2 x = 1 − sin2 x to transform the integral into one of the type described in Example ??. 2 Example 1.4. Z sin3 xdx Z Z sin2 x sin xdx = (1 − cos2 x) sin xdx Z Z = sin xdx − cos2 x sin xdx = = − cos x + Similarly one can solve Example 1.5. Z cos3 x sin2 xdx R cos3 x + C. 3 cos3 xdx. Z 2 Z cos x cos x sin xdx = (1 − sin2 x) cos x sin2 xdx Z Z = cos x sin2 xdx − cos x sin4 xdx 2 = sin3 x sin5 x − + C. 3 5 R R 1.6. In order to solve cos5 x sin3 xdx we express it first as cos5 x sin2 x sin x = RExample cos5 x(1 − cos2 x) sin xdx and then proceed as in the previous example. = One can use similar tricks to solve integrals which involve products of powers of sec x and tan x, by using Equation (2). Also, recall that the derivative of tan x is sec2 x while the derivative of sec x is sec x tan x. Example 1.7. Z tan5 x sec4 xdx Z 2 2 Z tan x sec x sec xdx = tan5 x(1 + tan2 x) sec2 xdx Z Z = tan5 x sec2 xdx + tan7 x sec2 xdx = = Example 1.8. Z tan3 x sec4 xdx 5 tan6 x tan8 x + + C. 6 8 Z = 4 Z tan x tan x sec xdx = tan x(sec2 x − 1) sec4 xdx Z Z = tan x sec x sec5 xdx − tan x sec x sec3 xdx = 2 sec6 x sec4 x − + C. 6 4 2 2 Trigonometric Substitutions R1√ One can easily deduce that 0 1 − x2 dx has value π4 . Why? Simply because √ the graph of the function y = 1 − x2 is√half a circumference of radius r = 1 (because if you square both sides of y = 1 − x2 you obtain x2 + y 2 = 1 which is the equation of a circle or radius r = 1). Therefore, the area under the graph is a quarter of the area of a circle. R1√ How does one compute 0 1 − x2 dx without using the geometry of the problem? This is the prototype of integral where a trigonometric substitution will work very nicely. Notice that neither substitution nor integration by parts will work appropriately. R1√ Example 2.1. Suppose we want to solve 0 1 − x2 dx with analytic methods. We will use a substitution x = sin θ (so θ will be our new pvariable of integration), √ because, as we know from Equation (1), 1 − x2 = 1 − sin2 θ = cos θ, thus getting rid of the pesky square root. Notice that dx = cos θdθ. We also need to find the new limits of integration with respect to the new variable of integration, namely θ. When x = 0 = sin θ we must have θ = 0. Similarly, when x = 1 = sin θ one has θ = π/2. We are now ready to integrate: Z 1 p 1 − x2 dx π/2 Z = π/2 Z cos2 θdθ (cos θ) cos θdθ = 0 0 0 π/2 Z = 0 1 1 + cos(2θ) dθ = 2 2 π/2 sin(2θ) = π/4. θ+ 2 0 Notice that we made use of Equation (4) in the second line. Rr√ Example 2.2. Similarly, one can solve 0 r2 − x2 dx by using a substitution p √ x = r sin θ. Indeed, r2 − x2 = r2 − r2 sin2 θ = r cos θ and dx = r cos θdθ. The limits of integration with respect to θ are again θ = 0 to θ = π/2 (check this!). Thus: Z r p r2 − x2 dx Z π/2 2 = 0 r (cos θ) cos θdθ = r 2 Z 0 = r2 π/2 cos2 θdθ 0 Z 0 π/2 r2 1 + cos(2θ) dθ = 2 2 θ+ sin(2θ) 2 π/2 = r2 π/4. 0 Thus, we have proved that a quarter of a circle of radius r has area r2 π/4 which implies that the area of such a circle is πr2 , as usual. √ substitutions usually work when expressions like r2 − x2 , √ The trigonometric √ r2 + x2 , x2 − r2 appear in the integral at hand, for some real number r. Here is a table of the suggested change of variables in each particular case: 3 If you see... √ 1 − x2 √ 2 2 √r − x 2 √1+x 2 2 √r + x 2 √x −1 x2 − r 2 try this... x = sin θ x = r sin θ x = tan θ x = r tan θ x = sec θ x = r sin θ p because... 2 p 1 − sin θ = cos θ 2 2 √r − sin θ = r cos θ 2 √ 1 + tan θ = sec θ 2 2 r + tan θ = r sec θ √ 2 √ sec θ − 1 = tan θ 2 sec θ − 1 = r tan θ Remark 2.3. The above are “suggested” substitutions, R √ they may not be the most ideal choice! For example, for the integral 2x 1 − x2 dx, the change u = 1 − x2 will work much better than x = sin θ. Example 2.4. We would like to find the value of Z 2 1 √ dx. √ 3 x2 − 1 2 x Since neither a u-substitution nor integration √ by parts seem appropriate, we try x = sec θ, dx = sec θ tan θdθ. When x = 2 = sec θ one has θ = π/4 while x = 2 implies θ = π/3. Hence: Z 2 √ 2 x3 √ 1 dx x2 − 1 Z π/3 = π/4 sec θ tan θ dθ = sec3 θ tan θ Z π/3 π/4 1 dθ = sec2 θ and the last integral is easy to compute using Equation (4). 4 Z π/3 π/4 cos2 θdθ