How does one find the indefinite integral of sine squared x.?
Q. I know that one can look up the indefinite integral of sine squared x - sin^2 (x) - on an integral table, but I need to actually find it myself for a Calc 1 assignment. I'm stumped, tried substituting u for sin(x) and also for sin^2(x), but to no avail. Please help.
Asked by Adinsx - Tue Nov 3 20:40:29 2009 - - 1 Answers - 0 Comments

A. Recall that sin^2(x) = (1 - cos(2x))/2 Taking the integral of that you should get (1/2)x - (1/4)sin(2x) + C.
Answered by unknown - Tue Nov 3 20:53:09 2009

Calculus 2 Integral Table Problem?
Q. Use the integral tables to determine: integral of 6x^2(sqrt(4-3x) dx And could you tell me what formula from the integral tables you are using please, thanks =) I know that this can be done very easily using mathematica, but that won't help me. I'm trying to understand the problem and just not get a simple answer, thanks!
Asked by MA35 - Mon Mar 24 18:20:42 2008 - - 2 Answers - 0 Comments

A. 6x^2* (4-3x) dx let 3x=4sin^2(u):sin^2(u)=3x/ 4: cos^2(u)=(4-3x)/4:cosu=(1 /2) (4-3x) 3dx = 8sinu cosu du dx = (8/3)sinucosu du now the integral becomes 6(16/9)sin^4(u)* (4-4sin^2(u) (8/3)sinu cosu 6*16*8/(9*3)sin^4(u)*2cos u sinu cosu 512/27 sin^5(u)*cos^2(u) du 512/27 sinu*sin^4(u)*cos^2(u) du 512/27 sinu*(1-cos^2(u))^2*cos^2 (u) du 512/27 sinu*(1+cos^4(u)-2cos^2(u ))*cos^2(u) du 512/27[ sinu cos^2(u)du + sinu cos^6(u)du - 2 sinu cos^4(u)du 512/27[-cos^3(u)/3 - cos^7(u)/7 + 2cos^5(u)/5] + c (512/27)cos^3(u)[-1/3 - (1/7)cos^4(u) + (2/5)cos^2(u)] + c substitute cosu = (1/2) (4-3x) (512/27)(1/8)(4-3x)^3/2[- 1/3 -(16+9x^2-24x)/112+(4-3x) /10] (64/27)(4-3x)^3/2[-560- 240 - 135x^2+360x+672- 504x]/1680 (64/27)(4-3x)^3/ [cont.]
Answered by mohanrao d - Mon Mar 24 18:55:32 2008

HOW did CRC math tables come up with the integral of sqrt(x^2-a^2)?
Q. i.e. can you derive it for me please if possible
Asked by cyran - Fri Dec 5 10:05:26 2008 - - 1 Answers - 0 Comments

A. let x = a secB then dx = a secB tanB dB sqrt(x^2 - a^2) dx = (a tanB) (a secB tanB dB) = a^2 secB tan^2(B) dB = a^2 secB (sec^2(B) - 1) dB = a^2 [ sec^3(B) dB - secB dB] = a^2 [1/2 secB tanB - 1/2 ln|secB + tanB|] + C now, secB = x/a thus tanB = (x^2-a^2)/a thus sqrt(x^2 - a^2) dx = a^2 [1/2 (x/a)( (x^2-a^2)/a) - 1/2 ln|x/a + (x^2-a^2)/a|] = x (x^2-a^2) / 2 - a^2/2 ln|x/a + (x^2-a^2)/a| + C
Answered by Alam Ko Iyan - Fri Dec 5 11:07:20 2008

integral table question 1/x^2 sqrt(4x^2 - 7)?
Q. ok. I need to find the integral of this using tables. S 1 / x^2sqrt(4x^2-7) dx S 1 / x^2 sqrt(4(x^2 - (7/4)) dx 1/2 S 1 / x^2 sqrt(x^2 - (7/4) dx my question is can i move dx to the top of the problem and make it dx/ x^2 sqrt(x^2 - (7/4)) i think that you can because then it will match one of my table integrals in the back. But i have to make sure. Thanks for your help!
Asked by TMT-AG - Wed Feb 18 20:54:38 2009 - - 1 Answers - 0 Comments

A. dx is ALWAYS at the top of a function and you do not need to move it. (Maybe it was not written correctly.)
Answered by unknown - Wed Feb 18 23:08:50 2009

Integral table?
Q. Does anyone know where to find a good integral table on the web? I would greatly appreciate it.
Asked by Jay - Tue Feb 5 17:26:29 2008 - - 1 Answers - 0 Comments

A. These are two pretty detailed ones: i use these all the time. I wish there was a poster I could get at a store that has the integrals on it - like one of those periodic tables for chemistry. Good question, though.
Answered by ivegotaquestion - Tue Feb 5 17:38:00 2008

What is the integral of x*Sqrt(x^2+2x+4)?
Q. I tried substitution to make it look like one on my integral tables but no luck.
Asked by Nate S - Tue May 22 09:46:54 2007 - - 4 Answers - 0 Comments

A. x^2+2x+4 = (x+2)^2 x*Sqrt(x^2+2x+4) dx = x * (x+2) dx = (x^2 + 2x) dx = (x^3 / 3) + x^2 + c c is a constant number
Answered by nelaq - Tue May 22 09:52:26 2007

integral of a quartic equation?
Q. I can't seem to find a certain integral in any of the integral tables integral of du / (a(u)^4+b(u)^2+c)^(1/2) I tried substituting u^2=v this gave me integral of dv / v^(1/2)*(av^2+bv+c)^(1/2) but that still isn't in the table. Any ideas? thanks
Asked by snoboarder2k6 - Wed Nov 11 22:55:47 2009 - - 1 Answers - 0 Comments

A. This will need huge explanations. Instead I am referring you to the following book : I am sure this will help you greatly.
Answered by Hemant - Sun Nov 15 07:21:33 2009

Solve integral 2/ sqrt(x^2 - 4x + 13) using table of integrals?
Q. I am trying to solve this using the tables found here A few hints would be appreciated.
Asked by my_craig1 - Wed Dec 19 00:49:29 2007 - - 3 Answers - 0 Comments

A. here is a hint: if you differentiate 2*asinh(1/3*x-2/3) you should get 2/sqrt(x^2 - 4x + 13) /m
Answered by perplexed* - Wed Dec 19 01:14:46 2007

Difficult integral. Need math expert!?
Q. Help me find the integral of 1/sqrt(600x + x^2) for a problem in my Dynamics class homework. I've looked at all the integral tables without any luck. Please help!
Asked by Jamin2112 - Tue Jan 13 00:43:19 2009 - - 4 Answers - 0 Comments

A. I = dx / (600x + x^2) I = dx / (600x + x^2 + 300^2 - 300^2) I = dx / [(x + 300)^2 - 300^2] --- let x+300 = 300 * sec p dx = 300 * sec p tan p dp I = 300 * sec p tan p dp / (300^2)(sec^2 p-1) I = sec p tan p dp /sec p I = tan p dp I = - ln |cos p| + C I = ln |sex p| + C replace (p) to (x) I = ln [(x+300)/300] + C
Answered by anil bakshi - Tue Jan 13 01:24:09 2009

Integral of 1/([cos^2](2y))?
Q. Okay I need to integrate 1 divided by consine squared 2y and none of my integral tables in my calc book really have anything. This is for a DiffEq class...the whole equation is 1/(cos[^2](2y)) dy = cos[^2](x) dx the right side of the equation is not an issue, but I do not know how to solve the left side... I do know the answer though which is 2tan(2y)-2x-sin(2x)=c Thanks
Asked by acardattack - Sun Jan 25 20:02:52 2009 - - 1 Answers - 0 Comments

A. if it is: 1/cos (2y) dy using the equation 1/cos x = 1 + tan x we have 1/cos (2y) dy = 1 + tan (2y) dy = (1/2) tan(2y) + c where c is constant. hope this helps.
Answered by Morteza - Mon Jan 26 17:22:21 2009

Integral of Inverse Trig Function?
Q. Calculate the integral arctan(x) dx. I know the answer because I can look it up in the table of integrals at the back of a calculus book, but I am drawing a blank trying to work it out.
Asked by Northstar - Sat Jan 3 23:04:41 2009 - - 4 Answers - 1 Comments

A. Integration by parts, followed by U-substitution. Since I know you're an expert I'll just not explain each step and do it. (arctan(x) dx) Let u = arctan(x). dv = dx du = 1/[1 + x^2] dx. v = x x arctan(x) - (x/[1 + x^2] dx ) x arctan(x) - (1/[1 + x^2] x dx ) Let u = 1 + x^2. du = 2x dx (1/2) du = x dx x arctan(x) - ( (1/u) (1/2) du ) x arctan(x) - (1/2) ( (1/u) du ) x arctan(x) - (1/2)ln|u| + C x arctan(x) - (1/2)ln|1 + x^2| + C x arctan(x) - (1/2)ln(1 + x^2) + C
Answered by Puggy - Sat Jan 3 23:16:34 2009

Determine the integral of ((4)/(21+8x-4x^2))dx by completingthe square, using substitution, table of integrals?
Q. Determine the integral of ((4)/(21+8x-4x^2))dx by completingthe square, using substitution, table of integrals?
Asked by J - Fri Oct 9 17:46:15 2009 - - 1 Answers - 0 Comments

A. Note that by completing the square: 21 + 8x - 4x^2 = 21 - 4(x^2 - 2x) = 21 - 4(x^2 - 2x + 1) + 4 = 25 - 4(x - 1)^2. Let u = x - 1, du = dx. Thus, we get integral 4 du/(25 - 4u^2). Now, use partial fractions: 4/(25 - 4u^2) = A/(5 + 2u) + B/(5 - 2u) ==> 4 = A(5 - 2u) + B(5 + 2u). u = 5/2 ==> 4 = 10B ==> B = 2/5. u = -5/2 ==> 4 = 10A ==> A = 2/5. Thus, we may rewrite the integral as integral (1/5) * [2/(5 + 2u) + 2/(5 - 2u)] du = (1/5) * (ln|5 + 2u| - ln|5 - 2u|) + C = (1/5) * ln|(5 + 2u) / (5 - 2u)| + C = (1/5) * ln|(5 + 2(x - 1)) / (5 - 2(x - 1))| + C. I hope this helps!
Answered by kb - Fri Oct 9 19:27:06 2009

Determine the integral of ((radical(2+(lnx)^2))/(x) )dx use substitution then table of integrals?
Q. Determine the integral of ((radical(2+(lnx)^2))/(x) )dx use substitution then table of integrals?
Asked by J - Fri Oct 9 17:42:31 2009 - - 1 Answers - 0 Comments

A. substitute u = ln(x) <=> x = e^(u) => dx = e^(u) du (2 + ln(x) )/x dx = ( (2 + u )/e^(u)) udu = (2 + u ) du Substitute u = 2 tan( ) => du = 2 (tan ( ) + 1) = 2 sec ( ) d (2 + u ) du = (2 + 2 tan ( )) 2 sec ( ) d = 2 (1 + tan ( )) sec ( ) d = 2 (sec ( )) sec ( ) d = 2 sec ( ) d From integral table [1]: sec ( ) d = sec ( ) sin( )/(n-1) + [(n-2)/(n-1)] sec ( ) d = sec ( ) tan( )/(n-1) + [(n-2)/(n-1)] sec ( ) d and sec ( ) d = ln|sec( ) + tan( )| 2 sec ( ) d = 2 ( sec( ) tan( )/(2) + [(1/(2)] sec( ) d ) = sec( ) tan( ) + ln|sec( ) + tan( )| + c = (1 + tan ( )) tan( ) + ln| (1 + tan ( )) + tan( )| + c substitute back tan( ) = u/ 2 (1 + tan ( )) tan( ) + ln| (1 + tan ( )) + tan( )| + c = (1 + [cont.]
Answered by schmiso - Tue Oct 13 07:47:20 2009

Integral( [1-(a/x)]^(1/2) dx ) ?
Q. Does anyone know the solution to this integral? I've been tackling it for a day now. I can't solve it using any method I've learned in two years of calc, and I've tried searching through tables of integrals with no success.
Asked by Dan B - Thu Oct 4 20:35:45 2007 - - 1 Answers - 0 Comments

A. Yikes! The square root with the "1 minus something" under it, try to think of it as a "1 minus something squared" instead. Then you might be able to pull off a trig substitution. [1 - (a/x)]^(1/2) = [1-(sqrt(a/x))^2]^(1/2). Now substitute sqrt(a/x) = sin (cos works too) Then -(1/2)x^(-3/2) * sqrt(a) dx = cos d -(1/(2a)) * (a/x)^(3/2) dx = cos d -(1/(2a)) * [sin^3 ] dx = cos d dx = -2a cos /[sin^3 ] d and sqrt[1-a/x] = sqrt[1 - sin^2 ] = cos . That should reduce your integral to a bunch of sines and cosines, which there are strategies for. Good luck.
Answered by K-Dub - Fri Oct 5 16:08:24 2007

How to solve this integral?
Q. This one is a little tricky, i tried using the table of common integrals but im not sure how to make it work! 1/ (3 + 2x - x^2) dx
Asked by elemental_funk - Mon Nov 2 22:32:33 2009 - - 1 Answers - 0 Comments

A. [1 / (3 + 2x - x )] dx = first complete the square inside the root by adding and subtracting 1: [1 / (3 + 1 - 1 + 2x - x )] dx = {1 / [4 - (x - 2x + 1)]} dx = {1 / [4 - (x - 1) ]} dx = substitute: x - 1 = 2sin d(x - 1) = d(2sin ) dx = 2cos d sin = (x - 1)/2 = arcsin[(x - 1)/2] yielding: {1 / [4 - (x - 1) ]} dx = {1 / [4 - (2sin ) ]} 2cos d = [1 / (4 - 4sin )] 2cos d = {1 / [4(1 - sin )]} 2cos d = (replacing 1 - sin with cos ) [1 / (4cos )] 2cos d = [1 /(2cos )] 2cos d = simplifying into: d = + C then substitute back arcsin[(x - 1)/2] for , concluding with: [1 / (3 + 2x - x )] dx = arcsin[(x - 1)/2] + C I hope it helps
Answered by germano - Tue Nov 3 19:30:16 2009

evaluate the integral from 0 to 2 of (51x^3)(sqrt(4x^2-x^4))dx ?
Q. Not really sure of how to do this and its on an exam in a week.. if anyone could walk me through it would appreciate it... you are supposed to use the table of integrals when solving it but i cant figure it out...
Asked by MISS_MARIE_010807 - Wed Apr 29 13:17:08 2009 - - 2 Answers - 0 Comments

A. 51x (4x - x^4) dx = first, rearrange a little the integrand as: 51x (4x - x^4) x dx = complete the square inside the root by adding and subtracting the missing term 4: 51x (4 - 4 + 4x - x^4) x dx = 51x [4 - (x^4 - 4x + 4)] x dx = 51x [4 - (x - 2) ] x dx = substitute (x - 2) = 2sin sin = [(x - 2)/2] = arcsin[(x - 2)/2] x = 2sin + 2 = 2(sin + 1) differentiate both sides: d(x ) = d[2(sin + 1)] 2x dx = 2cos d x dx = cos d yielding: 51x [4 - (x - 2) ] x dx = 51 [2(sin + 1)] [4 - (2sin ) ] cos d = 51 [2(sin + 1)] (4 - 4sin ) cos d = factor 4 out of the root: 102 (sin + 1) [4 (1 - sin )] cos d = 102 (sin + 1) [2 (1 - sin )] cos d = recall that 1 - sin =… [cont.]
Answered by germano - Wed Apr 29 23:32:53 2009

Integral of x=sqrt(3.68^2-(y-11.35)^2 ) - 4.52?
Q. I used the Table of Integrals for sqrt (a^2 - x^2) dx... but when i plug in the bound values, for some reason, the answer is incorrect. and i don't know why i have tried radians and degrees again and again it isn't right tho...
Asked by lil gUster - Fri Apr 10 22:02:55 2009 - - 1 Answers - 0 Comments

A. [2] (3.68 -(y-11.35) )-4.52 ([2] (3.68 -(y-11.35) )-4.52)dy= -2. 5 10 (-2000.0y+2. 27 10 ) (-1. 152801 10 -1000.0y +2. 27 10 y)+ 6. 7712arcsin(0.271739130 4y-3. 08423913)- 4. 52y
Answered by Joerg - Sat Apr 11 09:47:34 2009

How can I evaluate these without integration tables?
Q. I'll use "Int" for the indefinite integral symbol: Int[(tanx)^3(secx)^3]dx Int[(cos(3x))^4]dx
Asked by Richard H - Fri May 4 22:24:28 2007 - - 1 Answers - 0 Comments

A. Ok, Let's go with the first one : Int[(tanx)^3(secx)^3]dx Remember : tan^2x + 1 = sec^2x int[tanx*(sec^2x - 1)*sec*3x]dx int[tanx*(sec^5x - sec^3x)]dx remember : tanx = sinx / cosx secx = 1 / cosx int[(sinx / cos^6x)dx - int[sinx / cox^4x]dx Now we have easier integrals, remember : - d(cosx) = sinx -int[dcosx / cos^6x] + int[dcosx / cos^4x]dx 1/5*(cosx)^-5 - 1/3*(cosx)^-3 That's it Second problem : Int[(cos(3x))^4]dx Remember : 2*cos^2(3x) = cos6x + 1 cos^2(3x) = (cos6x + 1) / 2 cos^4(3x) = ( 1 + cos^2(6x) + 2*cos6x ) / 4 Let's replace it on the integral : int[ ( 1 + cos^2(6x) + 2*cos6x ) / 4)]dx 1/4*int[dx] + 1/4*int[cos^2(6x)] + 1/2*int(cos6x) 1/4*x + 1/12*sin(6x) + 1/4*int[cos^2(6x)]dx...(* *) This : 1/4*int[cos^2 [cont.]
Answered by anakin_louix - Fri May 4 22:34:30 2007

(math exams) Do you have to learn ALL the formulas of calculus or you'll get them in tables?
Q. I was wondering, cause I have managed to learn all the trigonometric formulae and it's really tough to learn those dervations and definite and indefinite integrals? So in general are you supposed to learn them>?
Asked by unknown - Sun Jan 18 05:02:41 2009 - - 2 Answers - 0 Comments

A. NO not at all. You have to read all the formulas of calculus. In the final exam you wil be given only with the log and antilog values and the sin cos and tan values of the angles. All the formulas should be memorised. Write the formulas in a sheet and daily go through those formulas to get used to it. During the final days you wont get those in mind. so start preparing from today.
Answered by Siva - Sun Jan 18 05:17:21 2009

How to use the integral rules from the table of integrals?
Q. 1. dx / (x^2 (4x^2+9) The Rule: du / u^2 (a^2 +u^2)=- (a^2+u^2 )/(a^2 u)+C 2. x (1+2x) dx The Rule: u (a+bu) du= 2/(15b^2 ) (3bu-2a)(a+bu)^3/2 3. sec^3 ( x) dx The Rule: sec^3 u du= 1/2 secu tanu +1/2 ln|secu tanu |+C
Asked by diamondzwdb_2006 - Fri Sep 26 17:26:02 2008 - - 1 Answers - 0 Comments