Chapter 13 TheTrigonometricFunctions - Purdue University Northwest
[Pages:4]Chapter 13 The Trigonometric Functions
13.3 Integrals Of The Trigonometric Functions
Integrals of trigonometric functions include:
sin x dx = - cos x + C sec2 x dx = tan x + C sec x tan x dx = sec x + C tan x dx = - ln | cos x| + C
cos x dx = sin x + C csc2 x dx = - cot x + C csc x cot x dx = - csc x + C cot x dx = ln | sin x| + C
Also, recall derivatives of trigonometric functions include:
Dx [sin x] = cos x Dx [cos x] = - sin x Dx [tan x] = sec2 x
Dx [csc x] = - cot x csc x Dx [sec x] = tan x sec x Dx [cot x] = - csc2 x
Exercise 13.3 (Integrals of Trigonometric Functions) 1. Find cos x dx.
From list above,
cos x dx =
(i) - ln | cos x| + C (ii) sin x + C (iii) sec x + C
2. Find sec x tan x dx.
251
252
Chapter 13. The Trigonometric Functions (LECTURE NOTES 14)
From list above,
sec x tan x dx =
(i) - ln | cos x| + C (ii) sin x + C (iii) sec x + C
3. Find tan x dx.
From list above,
tan x dx =
(i) - ln | cos x| + C (ii) sin x + C (iii) sec x + C
4. Find sin3 x cos x dx.
guess u = (i) sin x (ii) sin3 x (ii) 2x
then du = (i) cos x dx (ii) 2 sin x cos x dx (iii) 2 dx
recall, we are looking for the derivative here, use the trigonometric derivatives above
substituting u and du into f (x) dx,
sin3 x cos x dx = u3 du = 1 u3+1 + C = 3+1
(i)
u3 + C
(ii)
1 u4
4
+
C
(iii)
1 u2
2
+
C
but u = sin x, so
(i)
1 2
sin4
x
+
C
f (x) dx
=
1 u4 4
+C
=
(ii)
1 4
sin4 x
+
C
(iii)
1 3
sin4 x
+
C
5. Find tan3 x sec2 x dx. guess u = (i) tan x (ii) sec2 x (ii) 2x then du = (i) sec2 x dx (ii) 2 tan x sec2 x dx (iii) 2 dx
again, we are looking for the derivative here, use the trigonometric derivatives above
substituting u and du into f (x) dx, tan3 x sec2 x dx = u3 du = 1 u3+1 + C = 3+1
Section 3. Integration Of The Trigonometric Functions (LECTURE NOTES 14) 253
(i)
u3 + C
(ii)
1 u4
4
+
C
(iii)
1 u2
2
+
C
but u = tan x, so
(i)
1 4
tan2
x
+
C
f (x) dx
=
1 u4 4
+C
=
(ii)
1 4
tan3 x
+
C
(iii)
1 4
tan4 x +
C
6. Find sec2 5x dx.
guess u = (i) tan x (ii) sec2 x (iii) 5x
then du = (5x2-1) dx = (i) cos x dx (ii) sec2 x dx (iii) 5 dx
again, remember, derivative here
substituting u and du into f (x) dx,
sec2 5x dx =
sec2 5x
1 5
(5x) dx =
sec2 u
1 5
du =
(i)
-
1 5
tan
u
+
C
(ii)
1 5
csc u
+
C
(iii)
1 5
tan u
+
C
but u = 5x, so
f (x) dx =
(i)
1 5
tan 5x
+
C
(ii) tan 5x + C
(iii)
-
1 5
tan
5x
+
C
7. Find (sin x + x2)-1(cos x + 2x) dx.
guess u = (i) sin x + x2 (ii) cos x + 2x
then du = (cos x + 2x2-1) dx = (i) ln x dx (ii) (tan x sec x + 2x) dx (iii) cos x + 2x dx
again, remember, derivative here
substituting u and du into f (x) dx, sin x + x2 -1(cos x + 2x) dx = u-1 du =
(i) 2 ln |u| + C (ii) ln |u| + C (iii) 3 ln |u| + C
but u = sin x + x2, so
f (x) dx =
254
Chapter 13. The Trigonometric Functions (LECTURE NOTES 14)
(i) 2 ln | sin x + x2|+C (ii) ln | sin x + x2|+C (iii) 3 ln | sin x + x2|+C
8. Find sin3 (2x + x2) (2 + 2x) cos (2x + x2) dx.
guess u = (i) sin (2x + x2) (ii) (2x + x2) (iii) 2x
recall if u = f [g(x)] = sin (2x + x2) and g(x) = 2x + x2 and f (x) = sin x
and g(x) = (i) 2 + 2x (ii) 2x2 + 3x3 (iii) 1 + x and f (x) = (i) cos x (ii) - cos x (iii) ln x and so by chain rule
du = f [g(x)] ? g(x) = f 2x + x2 ? (2 + 2x) = cos(2x + x2) (2 + 2x) = dx
(i) cos(2x + x2) (ii) cos(2x + x2) (2 + 2x) (iii) (2 + 2x)
in other words du = (i) cos(2x + x2) (2 + 2x) dx (ii) (2 + 2x) dx
substituting u and du into f (x) dx,
sin3 2x + x2 (2 + 2x) cos 2x + x2 dx =
(i)
u3 du
=
1 u3+1
3+1
+C
=
4
u 4
+
C
(ii)
u2 du
=
1 u2+1
2+1
+C
=
3
u 3
+
C
(iii)
u du
=
1 u1+1
1+1
+C
=
2
u 2
+C
but u = sin (2x + x2), so
u4 f (x) dx = + C =
4
(i)
1 4
(sin
(2x
+
x2))4
+
C
(ii)
1 3
(sin
(2x
+
x2))3
+
C
(iii)
1 2
(sin
(2x
+
x2))2
+C
................
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