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Math 201-203-RE - Calculus II Integrals of Trigonometric Functions Page 1 of 11
What is the Antiderivative?
In a derivative problem, a function f(x) is given and you find the derivative f′(x) using the formulas and rules of
derivatives shown in a previous tutorial.
In an antiderivative problem, the derivative f′(x) is given and you find a function f(x) using the formulas
of antiderivatives shown later in this tutorial.
In other words, the antiderivative is the reverse question to the derivative question.
Trigonometric Antiderivatives Formulas
Z sin(x)dx = −cos(x)+C Z cos(x)dx = sin(x)+C
Z sec2(x)dx = tan(x)+C Z csc2(x)dx = −cot(x)+C
These formulas are verified if the derivative is applied to each antiderivative answer.
Sum, Difference & Division of terms
To find the antiderivative of a sum or a difference of terms, use appropriate trigonometric formulas for each term.
Z 2
Example 1: Determine 3 cos(x) −4 sin(x)+5 sec (x) dx
=3 sin(x) −4 −cos(x) +5 tan(x) +C ==3 sin(x)+4 cos(x)+5 tan(x)+C
Example 2: Determine Z 2 sec x−3 cos x dx −→ Rewrite: Z 2 sec x dx−Z 3 cos x dx
cos x cos x cos x
=Z 2sec x: sec xdx−Z 3(1)dx=Z 2 sec2 xdx−Z 3dx=2 tan x−3x+C
Example 3: Determine Z 4x sin x−5 csc x dx −→ Rewrite: Z 4x sin x dx−Z 5 csc x dx
sin x sin x sin x
=Z 4x(1)dx−Z 5csc x: csc xdx=Z 4xdx−Z 5 csc2 xdx
2 2
=2x −5(−cot x)+C =2x +5 cot x+C
Math 201-203-RE - Calculus II Integrals of Trigonometric Functions Page 2 of 11
Z 3 Z Z 3
Example 4: Determine 5−2sin xdx−→ Rewrite: 5 1 dx− 2 sin x dx
2 2 2
sin x sin x sin x
=Z 5csc2 xdx−Z 2 sin xdx=5 (−cot x)−2 (−cos x)+C =−5 cot x+2 cos x+C
Z 3cot x+4x sin x cos x Z cos x Z sin x cos x
Example 5: Determine dx −→ Rewrite: 3 sin x dx+ 4x dx
sin x cos x sin x cos x sin x cos x
=Z 3cos x: 1 dx+Z 4x(1)dx=Z 3 1 2 dx+Z 4xdx=Z 3csc2 xdx+Z 4xdx
sin x sin x cos x (sin x)
x2
2
=3(−cot x)+4 2 +C =−3 cot x+2x +C
Chain Rule Substitution
To find the antiderivative of a composite trigonometric function, use chain rule substitution and
appropriate trigonometric formulas.
Example 6: Determine Z 3 cos 4xdx = Z 3 cos(4x)dx
Let u = 4x (expression inside of trigonometric function is the angle)
du = 4dx (derivative of the inside expression)
the adjustment is necessary to replace into the original integral; we get: dx = 1 du
4
replace cos(4x) with cos(u) and dx with 1 du ; we get:
4
Z 3 cos(4x)dx −→ Z 3 cos(u) 1 du = Z 3 cos(u)du = 3 sin(u)+C (trigonometric formula is used)
4 4 4
Therefore, replace u = 4x to get answer:
Z 3 cos(4x)dx = 3 sin(4x)+C
4
Math 201-203-RE - Calculus II Integrals of Trigonometric Functions Page 3 of 11
Example 7: Determine Z 6 sin(2x+1)dx
Let u = 2x+1 (expression inside of trigonometric function is the angle)
du = 2dx (derivative of the inside expression)
the adjustment is necessary to replace into the original integral; we get: dx = 1 du
2
replace sin(2x+1) with sin(u) and dx with 1 du ; we get:
2
Z 6 sin(2x+1)dx −→Z 6 sin(u) 1 du = Z 3 sin(u)du = −3 cos(u)+C (trigonometric formula is used)
2
Therefore, replace u = 2x+1 to get answer: Z 6 sin(2x+1)dx = −3 cos(2x+1)+C
Example 8: Determine Z 4 sec2(8x+3)dx
Let u = 8x+3 (expression inside of trigonometric function is the angle)
du = 8dx (derivative of the inside expression)
the adjustment is necessary to replace into the original integral; we get: dx = 1 du
8
replace sec2(8x+3) with sec2(u) and dx with 1 du ; we get:
8
Z 4 sec2(8x+3)dx −→Z 4 sec2(u) 1 du = Z 1 sec2(u)du = 1 tan(u)+C (trigonometric formula is used)
8 2 2
Therefore, replace u = 8x+3 to get answer: Z 4 sec2(8x+3)dx = 1 tan(8x+3)+C
2
Z 2
Example 9: Determine 4x sin(x )dx
2
Let u = x (expression inside of trigonometric function is the angle)
du = 2xdx (derivative of the inside expression)
the adjustment is necessary to replace into the original integral; we get: 4xdx = 2du
replace sin(x2) with sin(u) and 4xdx with 2du ; we get:
Z 4x sin(x2)dx −→ Z sin(u)2du = Z 2 sin(u)du = −2 cos(u)+C (trigonometric formula is used)
2 Z 2 2
Therefore, replace u = x to get answer: 4x sin(x )dx = −2 cos(x )+C
Math 201-203-RE - Calculus II Integrals of Trigonometric Functions Page 4 of 11
Z x x
Example 10: Determine 3e cos(e +2) dx
x
Let u = e +2 (expression inside of trigonometric function is the angle)
x
du = e dx (derivative of the inside expression)
x
the adjustment is necessary to replace into the original integral; we get: 3e dx = 3du
x x
replace cos(e +2) with cos(u) and 3e dx with 3du ; we get:
Z x x Z Z
3e cos(e +2) dx−→ cos(u)3du = 3 cos(u)du = 3 sin(u)+C (trigonometric formula is used)
x Z x x x
Therefore, replace u = e +2 to get answer: 3e cos(e +2) dx=3 sin(e +2)+C
Z Z sin(x) Z −1
Tangent Formula Determine tan(x)dx = cos(x) dx = sin x: (cos x) dx
Let u = cos(x) (expression inside of power function)
du = −sin xdx (derivative of the inside expression)
the adjustment is necessary to replace into the original integral; we get: sin xdx = −du
replace cos x with u and sin xdx with −du ; we get:
Z tan(x)dx −→Z (u−1)(−1)du=Z −u−1du=−ln|u|+C (log formula is used)
Z
−1
Therefore, replace u = cos x to get answer: tan xdx = −ln|cos x|+C = ln (cos x) +C
−1 1
remember the ratio: (cos x) = cos x = sec x
Z tan xdx=−ln|cos x|+C =ln|sec x|+C
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