Use the given integral table to find and evaluate the given integral.
Given: ∫ 1/√6x - x2 dx
Transcribed Image Text: PRODUCTS OF TRIGONOMETRIC FUNCTIONS
sin(m + n)u
sin(m – n)u
+C
2(m – n)
cos(m - п)и
+C
cos(m + n)u
38.
sin mu sin nu du
40.
sin mu cos nu du
2(m + n)
2(m + n)
-u cos"+!
2(m – n)
sin(m +n)u , sin(m – n)u
+C
39. cos mu cos nu du:
| sin" u cos" u du
sin
m -1
- | sin"-² u cos" u du
41.
2(m + n)
2(т — п)
m + n
m +n
+! u cos"-! u
n-1
- sin" u cos"-² u du
m +n
m +n
PRODUCTS OF TRIGONOMETRIC AND EXPONENTIAL FUNCTIONS
42. " sin bu du :
a² + b²
(a sin bu – b cos bu) + C
43.
a" cos bu du =
a² + b²
(a cos bu + b sin bu) + C
POWERS OF u MULTIPLYING OR DIVIDING BASIC FUNCTIONS
51. ue" du = e" (u – 1) + C
Swe du =we
53. fu'a" du =
44. u sin u du = sin u – u cos u +C
Suco
u cos u du = cos u +u sin u + C
45.
52.
u"a"
|u° sin u du = 2u sin u + (2 – u²) cos u + C
u"-'a" du + C
In a
46.
In a
47.
u cos u du = 2u cos u + (u² – 2) sin u +C
e du
e du
54.
(n – 1)u“-1
n -
a" du
In a
a" du
* / u' cos u du
a"
is
n-1
48.
u" sin u du = -u" cos u + n
55.
(n – 1)un-I
du
In |In u| +C
u"
u" cos u du = u" sin u – n | u"-1 sin u du
56.u In u
49.
50. fu" lau
«" In u du
(n + 1)z [(n + 1) In u – 1] + C
POLYNOMIALS MULTIPLYING BASIC FUNCTIONS
57.
du
P'(u)e
[signs alternate: + - + -..]
|p(u) sin au du = -- p(u) cos au +p'(u) sinau +
| P(u) cos au du = - p(u) sin au + p'(u) cos au -
P" (u) cos au
[signs alternate in pairs after first term: ++-- ++--...]
58.
P"(u) sin au –
[signs alternate in pairs: ++ -- ++--..]
59.
Transcribed Image Text: TABLE OF INTEGRALS
BASIC FUNCTIONS
Sa" du = na
1.
du =
n+ 1
10.
+C
du
|" = In lu| +C
In u du %3D u In u —и+С
2.
11.
|" du – " +C
12. cot u du = In |sin u| + C
3.
| sec u du = In |sec u + tan u| + C
= In (tan (7 + ju) | +c
4.
sin u du = - cos u + C
13.
5.
cos u du = sin u +C
| csc u du = In |csc u – cot u| +C
= In |tan {u| + C
14.
6.
tan u du = In |sec u|+C
lu = u sin=' u + V1 – u² +C
cot-' u du = u cot-' u + In ī+u² +C_
15.
cos- u du = u cos¯-lu – V1 - u² +C
sec-" u du = u sec-u – In \u + vu² – 1| +C
8.
16.
tanu du = u tan-u – In V1 + u² + C
- | csc-" u du = u csc-" u + In [u + /u? – 1+C
17.
9.
RECIPROCALS OF BASIC FUNCTIONS
18.
du = tan u F sec u +C
22.
du = }(u # In [sin u ± cos u|) + C
1± sinu
1t cot u
19.
du = - cot u ± csc u + C
23.
du = u + cot u F csc u + C
1t cos u
1t sec u
20. Tt tan u
du = }(u± In |cos u± sin u|) + C
24. It csc u
du = u – tan u ± sec u + C
21.
du = In [tan u| + C
du = u – In(1 ± e") + C
25.
sin u cos u
POWERS OF TRIGONOMETRIC FUNCTIONS
· | sin² u du = }u – } sin 2u + C
| cot? u du = - cot u – u + C
26.
32.
27. | cos² u du = ļu + į sin 2u + C
33. |
sec²
= tan u + C
28. | tan? u du = tan u – u +C
34. csc² u du = – cot u + C
1
sin" u du = -- sin"
u cos u + "=' | sin"-2 u du
cot" u du :
cot"-1
: / cor"-2,
29.
35.
и du
n- 1
30. f
n -
u sin u +
cos"-2 u du
п — 2
i/ sec"-2
cos" u du = - cos"
36.
sec" u du =
sec"-2 u tan u +
и du
n- 1
n - 2
tan" u du =
| csc" u du
csc"-2 u cot u +
| csc"-2 u du
31.
tan"
и du
37.
tan
n- 1
n-1
n- 1
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 2 images