/ 2 ___ | x V x | ------- dx | 1/3 / x |
/ 13/6 6 19:6 = | x dx = -- x + C / 19
/ = | ln4(x) | ------ dx / 5 x |
say ln(x) = u , then du = (1/x) dx / 4 5 5 = | u u ln (x) | ----- du = ---- + C = ------ + C / 5 25 25
/ |sqrt(3x-7)dx / |
1 / 2 = - |sqrt(3x-7)d(3x-7) = - (3 x - 7)3/2 +C 3 / 9
/
| x2
= | --------- dx
| _______
/ | 6
\| 9 - x
|
say x3 = t then 3x2dx = dt
3
/ 1 dt 1 t 1 x
= | - --------- = - arcsin(-) = - arcsin(---) + C
| _______ 3 3 3 3
| | 2
/ 3 \| 9 - t
/ | |x2.tan(x3+ 1) dx | / |
say x3+ 1 = u then du = 3 x2dx
/ /
1 | 1 | sin(u)
=- | tan(u) du = - | -------du
3 | 3 | cos(u)
/ /
say cos(u) = t then dt = -sin(u)du
/
1 | dt 1 1
=- - | --- = - - ln|t| +C = - - ln|cos(u)| +C
3 | t 3 3
/
1
= - - ln|cos(x3 + 1)| +C
3
/
| 2x + arctan(x)
|----------------dx
| 1 + x2
/
|
/ /
| 2 x | arctan(x)
= |-------------dx + |-------------dx
| 1 + x.x | 1 + x2
/ /
In the first part say 1 + x2 = t , then 2xdx = dt
In the second part say arctan(x) = u , then du = dx / (1+x2)
/ dt /
= | --- + | u du
/ t /
2 2
u 2 arctan (x)
= ln|t| + ---- = ln|1+x | + ----------- + C
2 2
/ ax | e | --------dx | ax / e + b |
say eax + b = u then aeaxdx = du 1 / du 1 1 ax = - | --- = - ln|u| = --- ln|e + b| + C a / u a a
/ I= | (x + 3) cos(x) dx / |
say (x + 3) = u and cos(x) dx = dv
Integration by parts gives
/
I = (x + 3) sin(x) - | sin(x) dx
/
= (x + 3) sin(x) + cos(x) + C
/ | 2 + x | -------- dx | 2 - x / |
/ / /
| x + 2 | x -2 + 4 | 4
= - | -------- dx = - | --------- dx = - | (1 + -------) dx
| x - 2 | x - 2 | x - 2
/ / /
= -x -4 ln |x - 2| + C
/ | x arcsin(x) | ----------- dx | ________ | | 2 | \| 1 - x / |
say arcsin(x) = u and the rest = dv
then
/ ________
| x | 2
v = | ----------- dx = ... = - \| 1 - x
| ________
| | 2
| \| 1 - x
/
Integration by parts gives
________ / ________
| 2 | | 2 1
= - arcsin(x) \| 1 - x + | \| 1 - x . --------------dx
| ________
/ | 2
\| 1 - x
________
| 2
= - arcsin(x) \| 1 - x + x + C
/ | |ex . cos( 1 + ex) dx | / |
say 1 + ex = u then exdx = du / = | cos(u) du = sin u + C = sin ( 1 + e )x + C /
Find
________
/5 | 2
| \| x - 9
I = | ---------- dx
| x
/3
|
First we calculate a primitive function
________ ________
/ | 2 / | 2
| \| x - 9 | x\| x - 9
| ---------- dx = | ---------- dx
| x | 2
/ / x
say x2 - 9 = u2 , then x dx = u du
/ / 2 /
| u.u du | u + 9 -9 | du
= | ---------- = = | ---------- du = u - 9 | ----------
| 2 | 2 | 2
/ u + 9 / u + 9 / u + 9
= u - 3 arctan(u/3)
________
________ | 2
| 2 \| x - 9
= \| x - 9 - 3 arctan ----------
3
________ |5
________ | 2 |
| 2 \| x - 9 | 4
Thus, I = \| x - 9 - 3 arctan ---------- | = 4 - 3 arctan(---) = 1.218
3 | 3
|3
Find
/pi/2
| 1 - cos(u)
g(x) = | -------------- du
| sin2 (u)
/ x
|
Let 2t = u ; then du = 2 dt and first we'll calculate
/
| 1 - cos(2t)
2 |-------------- dt =
| sin2 (2t)
/
With carnot we write 1 - cos(2t) = 2 sin2(t)
and sin(2t) = 2 sin(t) cos(t) => sin2 (2t) = 4 sin2(t) cos2(t)
So, the indefinite integral is equal to
/
| 2 sin2(t)
2 |----------------------- dt =
| 4 sin2(t) cos2(t)
/
/
| 1
|---------------- dt = tan(t) + C = tan(u/2) + C
| cos2(t)
/
From this result, we write
|pi/2
g(x) = tan(u/2) | = tan(pi/4) - tan(x/2)
|x