..READ THIS FIRST..Problems
about integration
READ THIS
FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
Problems about integration
Level 1 problems
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/ 2 ___ | x V x | ------- dx | 1/3 / x
/ 13/6 6 19:6 = | x dx = -- x + C / 19
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/ = | 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
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/ |sqrt(3x-7)dx /
1 / 2 = - |sqrt(3x-7)d(3x-7) = - (3 x - 7)3/2 +C 3 / 9
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/ | x2 = | --------- dx | _______ / | 6 \| 9 - xsay 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
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/ 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 /
Level 3 problems
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Find ________ /5 | 2 | \| x - 9 I = | ---------- dx | x /3First 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) / xLet 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)
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