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Section 5 – 1D: Integrals involving the basic Trigonometric Functions The integral of a function is the anti derivative of that function. If you know the derivative of a function F(x) is f (x) then you also know that the integral of ∫ f (x) dx = F(x) . We know the derivatives or the 6 basic trigonometric functions. This allows us to state the derivatives of these 6 functions as an integral. Example 1 Example 2 Example 3 We know that d sin(x) = cos(x) dx We know that d sec(x) = sec(x)tan(x) dx We know that d tan(x) = sec 2 (x) dx so we know the integral so we know the integral so we know the integral ∫ cos(x) dx must be sin(x) + C ∫ sec(x) tan(x) dx must be sec(x) + C 2 ∫ sec (x) dx must be tan( x) + C ∫ cos(x) dx = sin(x) + C ∫ sec(x) tan(x) = sec(x) + C 2 ∫ sec (x) = tan( x) + C Example 4 Example 5 Example 6 We know that d cos(x) = − sin(x) dx We know that d csc(x) = − csc(x)cot(x) dx We know that d cot(x) = − csc 2 (x) dx so we know the integral so we know the integral ∫ − sin(x) dx must be cos(x) + C so we know the integral − csc(x) cot( x) dx must be csc(x) + C ∫ − sin(x) dx = cos(x) + C ∫ − csc(x)cot(x) dx = csc(x) + C 2 ∫ − csc (x) = cot( x) + C 2 ∫ − csc (x) dx must be cot(x) + C Note: The first 3 integrals are commonly used in this course. The use of the last 3 is less common but we state all 6 for completeness. Examples 3 , 4 , and 5 above are all functions with a coefficient of negative 1. ∫ − sin(x) dx = cos(x) + C ∫ − csc(x)cot(x) dx = csc(x) + C 2 ∫ − csc (x) = cot( x) + C We will now develop techniques to integrate these functions with a coefficient of positive 1. ∫ sin(x) dx Section 5 – 1D ∫ csc(x)cot(x) dx Page 1 2 ∫ csc (x) © 2015 Eitel An Integral can be multiplied by a constant times its reciprocal 1 ∫ f (x) dx = • ∫ a• f (x) dx a 1 because • a = 1 a so ∫ f (x) dx = − ∫ − f (x) dx because − 1• −1 = 1 This rule that allows us to multiply the outside and inside of an integral by –1 ∫ f (x) dx = − ∫ − f (x) dx Example 7 Example 8 Example 9 d cos(x) = − sin(x) dx Find d csc(x) = − csc(x)cot( x) dx Find d cot(x) = − csc 2 (x) dx Find ∫ sin(x) dx ∫ csc(x)cot(x) dx 2 ∫ csc (x) dx multiply inside and outside by − 1 because − 1• −1 = 1 to get − ∫ − sin(x) dx multiply inside and outside by − 1 because − 1• −1 = 1 to get − ∫ − csc(x) cot( x) dx multiply inside and outside by − 1 because − 1• −1 = 1 ∫ sin(x) dx = − cos(x) + C ∫ csc(x)cot(x) dx = − csc(x) + C to get − ∫ − csc 2 (x) dx 2 ∫ csc (x) dx = − cot(x) + C These 3 integral rules above match the rules commonly listed for the basic trig integrals. The listed Trigonometric integrals rules The 6 rules for integrals shown below require the function to be exactly the Trigonometric function shown with the variable be exactly x. ∫ cosx dx = sin x + C ∫ sec x tan x dx = sec x + C 2 ∫ sec x dx = tan x + C ∫ sinx dx = − cos x + C ∫ csc x cot x dx = − csc x + C 2 ∫ csc x dx = − cot x + C THESE RULES CANNOT be used for functions like ∫ cos3x dx or ∫ sinx 2 dx or ∫ sec2 (x + 2) dx because the variable is not exactly x. Section 5 – 1D Page 2 © 2015 Eitel What's the difference between ∫ sin(x) dx and ∫ − sin(x) dx The derivative of cos(x) yields an expression with a negative coefficient. d cos(x) = − sin(x) dx Since we are determining anti derivatives we now have a rule for the integral of ∫ − sin(x) d cos(x) = − sin(x) dx so ∫ − sin(x) dx = cos x + C The use of the ∫ f (x) dx = − ∫ − f (x) dx rule for integrals allows us to develop a rule for ∫ sin(x) dx ∫ sin(x) dx multiply inside and outside by − 1 to get − ∫ − sin(x) dx = − cos(x) + C When we start to create a list or table of integral rules for basic functions you would think we would would use ∫ − sin(x) dx = cos x + C as it is derived from an anti derivative That is not the common choice. All tables of basic integrals list the function with a positive coefficient. The student then memorizes that version and uses it as the basic rule. This causes 3 of the integrals for basic trigonometric functions to be written as shown below ∫ sin(x) dx = − cos x + C is listed as the basic integral instead of ∫ − sin(x) dx = cos x + C 2 ∫ csc x dx = − cot x + C is listed as the basic integral instead of 2 ∫ − csc x dx = cot x + C ∫ csc x cot x dx = − csc x + C is listed as the basic integral instead of ∫ − csc x cot x dx = csc x + C Section 5 – 1D Page 3 © 2015 Eitel Useful Trigonometric Identities sec x = 1 cos x sin2 x + cos2 x = 1 csc x = 1 sin x tan x = tan2 x +1 = sec 2 x 1 cot x cot 2 x +1 = csc 2 x cos2x = cos2 x − sin 2 = 2 cos2 x − 1 = 1− 2 sin2 x sin2x = 2 sin x cos x Integrals Involving Trigonometric Identities Example 1 ∫ sin 2x dx cos x the functions are not both in terms of x but you can use the double angle rule sin2x = 2 sin x cos x Example 2 1− sin2 x dx ∫ cos x you can use sin2 x + cos2 x = 1 to derive 1− sin2 x = cos2 x cos2 x dx cos x 2sinx cos x dx ∫ cos x ∫ = 2∫ sin x dx which integrates to = − 2cos x + C = ∫ cos x dx which integrates to = sinx + C Example 3 Example 4 ∫ sinx(csc x − 1) dx ∫ tanθ (cot θ − sec θ ) dθ ∫ (sin x csc x − sin x ) dx ∫ (tan θ cot θ − sec θ tanθ ) dθ ⎛ ∫ ⎝ sin x ⎛ ∫ ⎝ tanθ 1 ⎞ − sin x dx ⎠ sin x 1 ⎞ − sec x tan x dθ ⎠ cot x = ∫ 1 dx − ∫ sin x dx = ∫ 1 dθ − ∫ secθ tanθ dθ which integrates to = x + cos x + C which integrates to = θ − sec θ + C Section 5 – 1D Page 4 © 2015 Eitel