Integration of sinx cosx/a^2sin^2x b^2 cos^2x

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Integration of sinx cosx/a^2sin^2x b^2 cos^2x

Dear Student, Please find below the solution to the asked query : Let I = sin x . cos xa2sin2x + b2cos2x dx=sin x . cos xa2sin2x + b2-b2sin2x dx=sin x . cos xa2-b2 sin2x + b2 dxput sin2x = t2 sin x . cos x dx = dtsin x . cos x dx = dt2Now, I = 12dta2-b2t2 + b2=12a2-b2 dtt2 + b2a2-b2=12a2-b2dtt2 + ba2-b22=12a2-b2?a2-b2b tan-1ta2b2b + C=12b?1a2-b2 tan-1a2-b2 sin2xb + C Do let us know in case of any further concerns. Regards 1/a? log|cosex-cotx| + 1/b? log|secx + tanx | + C 1/2 sin2x / a?sin?x + b?cos?x dx let , a?sin?x + b?cos?x = t dx = dt / sin2x (a?-b?) 1/2(a?-b?) log|a?sin?x + b?cos?x| + C , I noted down Wrong question in the above answer :P sorry for that this is correct now.. Something went wrong. Wait a moment and try again. Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,. Buy Now Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,. Buy Now Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,. Buy Now Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,. Buy Now Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,. Buy Now find the volume bounded by the parabolic cylinder z=4-x^2 and the planes x=0, y=0, y=6 and z=0 I just want some help figuring out the limits of this question : So for z, we have to integrate from z=0 to z=4-x^2 (am i right?) W.r.t. y, we should integrate Find the area of the region enclosed between y = 2sin (x) and y = 2cos (x) from x = 0 to x = pi/4. Thanks for your help :) Find the length of the curve correct to four decimal places. (Use a calculator to approximate the integral). r(t) = (cos t, 2t, sin 2t), from (1, 0, 0) to (1, 4, 0) This what I did. r'(t)=-sin(t),2,2cos(2t) Find the particular solution to y ' = 2sin(x) given the general solution is y = C - 2cos(x) and the initial condition y(pi/2) = 1 A weight is attached to a spring that is oscillating up and down. it takes 3 sec. for the spring to complete one cycle, and the distance from the highest to lowest point is 4 in. What equation models the position of the weight at time t seconds? a. Multiply then use fundamental identities to simplify the expression below and determine which of the following is not equivalent (2 - 2cos x)(2 + 2cos x) a.4 - 4cos^2 x b . 4 - cos^2 x c.4/(1 + cot^2 x) d. 4sin^2 x e.4/(csc^2 x) 1) what is the phase shift of f(x) = -2sin(3xpi)+1 2) What is the period of f(x) = -2sin(3x-pi)+1 Consider the function f(x) whose second derivative is f(x)=10x+2sin(x). If f(0)=4 and f'(0)=4, what is f(x)? Please do not include the constant (+C) in your answer. I got f(x) to be (10/6)x^3-2sin(x), but don't really know what to do with the f(0)=4 if y=3e^(2x)cos(2x-3) verify that d^2y/dx^24dy/dx+8y=0 plz help me i tried all i could but it become too complicated for me here set u=3e^(2x) v=cos(2x-3) du/dx=6e^(2x) i used chain rule dv/dx=-2sin(2x-3) dy/dx=-3e^(2x)sin(2x-3)+cos(2x-3)6e^(2x) d^2y/dx^2 Solve each equation for 0 is less than and equal to "x" is less than and equal to 360 3sinx = 2cos^{2}x ----- I don't know how to solve this equation...this is what I have, but I don't know if I'm on the right track or not 3sinx = 1 - 2sin^{2}x 3sinx - 1 + Use the formula in Exercise 42 to find the curvature. x=acos(t), y=bsin(t) Exercise 42 formula: k = l(x' y'' - y' x'')l / (x'^2 + y'^2)^(3/2) This is what I did but don't know why my answer not correct. x'=-asin(t), y'=bcos(t),x''=-a^2cos(t), Calculate the area of the common interior of r = 2sin(theta) and r = 2cos(theta) simplify 2sin(pi/6)2cos(pi/6) to an expression of the form (a sin(theta)) Use the Pythagorean identity to show that the double angle formula for cosine can be written as a) cos2x = 1 - 2sin^2x b) cos2x = 2cos^2x - 1 Given that 2cos^2 x 3sin x -3 = 0, show that 2sin^2 x + 3sin x +1 = 0? Hence solve 2cos^2 x + 4sin x - 3 = 0, giving all solutions in the interval 0 x 2 A circle is defined by the parametric equations x = 2cos(2t) and y = 2sin(2t) for t is all real numbers. (a) Find the coordinates of the point P on the circle when t = (4*pi)/3. (b) Find the equation of the tangent to the circle at P. Can you please show Find all solutions on the interval [0.2pi) A) -3sin(t)=15cos(t)sin(t) I have no clue... b) 8cos^2(t)=3-2cos(t) All i did was move around the equation to make an quadratic for B. so -8cos^2(t)-2cos(t)+3 = 0 Integrate: f (x)/(2x + 1) dx let f represent integrate sign let u = x, du = dx => dx = du = f (u)/(2u + 1) du = (2u + 1)^(-1) du = (1/2)u^2 (ln|2u + 1|) + c = (1/2)x^2 (ln|2x + 1|) + c ...what did I do wrong? The correct answer is (1/2)x - (1/4)ln|2x + 1| how does thederivative of -2xcos(x^2)= -4xcos(x^2) -2sin(x^2) and not -4x^2cos(x^2) -2sin(x^2)? given x=2cos(t)+t*sin(t) and y=2sin(t)-t*cost(t), with t:[pi, 2pi] find all critical values. (4 answers) Verify that each of the following is an identity. tan^2x-sin^2x=tan^2xsin^2x I can get it down to cos^2 on the right, but cannot get it to work out on the left. secx/cosx - tanx/cotx=1 On the left I got down to 1-tan^2, but that clearly doesn't equal 1.... Find circumference of the circle r=2acos theta. s= Int (0 to 2pi) of Sqrt(4a^2cos^2 theta+4a^2sin^2 theta)d theta =Int (0 to 2pi)2a*Int theta d theta =2a(2pi-0)=4a*pi Book shows 2a*pi. Am I wrong somewhere? Integrate:Sinxcosx/a^2cos^2(x)+b^2sin^2(x))dx??? plz i need the full working too hard for me find the exact solutions 2cos^2x+3sinx=0 the way it stands, that is a "nasty" question. Are you sure the second term isn't 2sin(2x) ? no, its as i wrote it. then it's got me stymied, I must be missing something rather obvious, sorry! thanks for trying! Verify the identities. Cos^2x - sin^2x = 2cos^2x - 1 When verifying identities, can I work on both side? Ex. 1 - sin^2x - sin^2x = 1 - 2sin^2x 1 - 2sin^2x = 1 - 2sin^2x y=3e^(2x)cos(2x-3) verify that d^2y/dx^2-4dy/dx+8y=0 plz help me i tried all i could but it become too complicated for me here set u=3e^(2x) v=cos(2x-3) du/dx=6e^(2x) i used chain rule dv/dx=-2sin(2x-3) dy/dx=-3e^(2x)sin(2x-3)+cos(2x-3)6e^(2x) d^2y/dx^2 2cos^2a*cos^2b+2sin^2asin^2b-1=cos2acos2b If f(x)=sin(2x), find f"(x) A. 2cos(2x) B. -4sin(2x) C. -2sin(2x) D. -4sinx E. None of these Is it B from using chain rules? How do I find the unit tangent vector with what I have x =2cos(t) , y=2sin(t), z =2t The formula is T(t)=r'(t)/lr'(t)l. Show steps so I know. Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. x=4+2cos(theta) y=-1+2sin(theta) Prove: sin2x / 1 - cos2x = cotx My Attempt: LS: = 2sinxcosx / - 1 - (1 - 2sin^2x) = 2sinxcosx / - 1 + 2sin^2x = cosx / sinx - 1 = cosx / sinx - 1/1 = cosx / sinx - sinx / sinx -- Prove: 2sin(x+y)sin(x-y) = cos2y - cos2x My Attempt: RS: = 1 - 2sin^2y - 1 - given x=2cos(t)+t*sin(t) and y=2sin(t)-t*cost(t), with t:[pi, 2pi] find all critical values. by finding dy/dx and setting it equal to zero, i only find one value, t=6.121, within that domain. that gives you x=.987. I'm suppose to get three other answers. cosA= 5/9 find cos1/2A are you familiar with the half-angle formulas? the one I would use here is cos A = 2cos^2 (1/2)A - 1 5/9 + 1 =2cos^2 (1/2)A 14/9 =2cos^2 (1/2)A cos (1/2)A = (7)/3 (cosx - sinx)^2 + (cosx + sinx)^2 = 2 Iam in this step: 2cos^2(x) + 2sin^2(x) = 2 How can I make the equation on the right equal 2 integrate (sin^8 x - cos^8 x)/(1 - 2sin^2 x * cos^2 x) w.r.t. x Use the sum or difference identity to find the exact value of sin255 degrees. My answer: (-sqrt(2)- sqrt(6)) / (4) Find the value of tan (alpha-beta), if cos alpha= -3/5, sin beta= 5/13, 90 Find sin(x/2) if sin(x)= -0.4 and 3pi/2 < or equal to (x) < or equal to 2pi Let's use cos 2A = 1 - 2sin2 A and we can match cos x = 1 - 2sin2 (x/2) so we will need cos x we know sin x = -.4 and x is in the fourth quadrant, so the cosine will be positive. If 2cos(x)-1 = (1+3)/2 and 2cos(x)+1 = (1-3)/2, find the value of cos4x. the problem is 2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is what i've done so far: 2cos^2x+sinx-1=0 2cos^2x-1+sinx=0 cos2x + sinx =0 1 - 2sin^2x + sinx = 0 -2sin^2x+sinx-1=0 Question: Prove that [integrate {x*sin2x*sin[/2*cos x]} dx] /(2x-) } from (0-) = [ integrate {sin x*cos x*sin[/2*cos x} dx ] from (0-). My thoughts on the question: We know that integrate f(x) dx from (0-a) = integrate f(a-x) dx from (0-a) From ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side and try to change it to How do I do the cross product of the following (/8 cross(/ 4)? Which expression is equivalent to sin(3x) + sin x? A) 2cos(2x)sin x B) 2sin(2x)sin x C) -2sin(2x)cos x D) 2sin(2x)cos x E) -2cos(2x)sin x Which expression is equivalent to sin(3x) + sin x? A. 2cos(2x)sin x B. 2sin(2x)sin x C. -2sin(2x)cos x D. 2sin(2x)cos x E. -2cos(2x)sin x I am given a vector function (in other words) : (2cos(t) + cos(2t))i + (2sin(t) + sin(2t)j + 0k Compute the slope dy/dx and concavity d^2x/(dx^2) at t = pi/3. I understand that finding the slope of a vector is possible by using dy/dx such that I have to find an exact solutions on the interval [0,pi) for both of these problems-- 1) 2sin2x = sqrtx and 2) 1-2sqrt2 sinxcosx = 0 I know this much: 1) sin2x = (sqrt x)/2 = 2sinxcosx = (sqrt x)/2 What do I need to do next? 2) 1/(2 sqrt2) = sinxcosx = how does 2cos^2 x - 2sin^2 x = 2cos2x Evaluate Each Expression: a) 2cos^2 2/3 - 1 b) cos^2 7/8 - sin^2 7/8 c)2sin 11/12 x cos 11/12 d) 1 - 2sin^2 /2 2cos^2a*cos^2b+2sin^asin^b-1=cos2acos2b how to prove it 2cos^2a*cos^2b+2sin^2asin^2b-1=cos2acos2b how to prove it,please help me Simplify (double angle formulas) a) cos3xcos2x + sin3xsin2x b) 2cos^2 pi/6 - 1 c) 1 - 2sin^2 pi/12 how would you find the angle to (3)^.5sin(2x)-2cos(2x)+2sin^2(x)=1 I know that you have to use the trig identities but then when i substitue the sin and cos(2x) in i don't know what to do. please help Please do help solve the followings 1) Integrate e^4 dx 2) Integrate dx/sqrt(90^2-4x^2) 3) Integrate (e^x+x)^2(e^x+1) dx 4) Integrate xe^x2 dx e^4 is a constant. 3) let u= e^x + x du= (e^x + 1)dx 4) let u= x du=dx v= e^x dv= e^x dx By using appropriate software/programming, sketch the following vector functions: 1) 2cos(2t)i + 2sin(3t)j for t within o to 2pai 2) 16sin^3ti + (13cost - 5cos2t 2cos3t - cos4t)j for t within 0 to 2pai 1) Integrate Cos^n(x) dx 2) Integrate e^(ax)Sinbx dx 3) Integrate (5xCos3x) dx I Will be happy to critique your thinking on these. 1) Derive a recursive relation. 2) Simplest by replacing sin(bx) by Exp[i b x] and taking imaginary part at the end. 3) First Given f(x)=sin(x)-2cos(x) on the interval [0,2pi]. a) Determine where the function is increasing and decreasing. b) Determine where the function is concave up and concave down. ________________________________________ I know: f(x)=cos(x)+2sin(x) Use implicit differentiation to find dy/dx. e^4x = sin(x+2y). This is a practice problem. It says the correct answer is 4e^x/(2sin(x+2y)) but I keep getting 4e^(4x)/(2cos(x+2y)). I thought the derivative of e^(4x) would be 4e^(4x), not 4e^(x). And I use Identify the amplitude, period, phase shift and vertical shift when appropriate. 1. y=sin(x+pi/2) 2. y=sin (x -pi) ? 2 3. y=1/2cos(2x) 4. y=2sin(2x+pi)-3 find the maximum value of f(x)= 3cos(x)-2sin(x) I know you have to take the derivative but after the derivative I don't know how to solve it. f'(x)=-3sin(x)-2cos(x)=0 I don't know what to do after this please help should i use substitution?? if yes how should should i use it? plz i need some directions? k plz someone?...so far i used trig. substitution. i got a=8, so i used x=asin(?)so according to this substitution i got x=8sin(?) and dx=8cos(?) d?...then i 2Sin tita+2Cos tita=1.5.find tita Your CEO is concerned that too much productivity is lost by having employees call each other, only to be directed to voice mail. He asks you if something can be done to counter this. What do you suggest? a) Integrate Dropbox into the company network. b) Question : Integrate [x/(1+(sin a*sin x))] from 0 to pi My first thought was to apply integrate f(x) dx= f(a-x) dx method Which simplified the integral into; 2I = integrate [pi/(1+(sin a*sin x))] dx , cancelling out x Then I made the integral into the form Given the curve C in parametric form: C : x =2cos(t) , y=2sin(t), z =2t; 0t2 Find each of the following in terms of t : a) The velocity v(t) b) The speed ds/dt c)The acceleration a(t) d) The unit tangent vector T(t) e)The curvature k and the Question : The length of time(in hundreds The length of time(in hundreds of hours) for the failure of a transistor is a random variable Y with distribution function ; F(Y) = { 0 : if y=0 } Find the probability that the transistor operates for at least 200 1. 3cot^2 (x) - 1 = 0 My answer: pi/3, 2pi/3, 4pi/3, 5pi/3 2. 4cos^2 (x) - 1 = 0 My answer: pi/3, 2pi/3, 5pi/3, 4pi/3 3. 2sin (x) + csc (x) = 0 My answer: unknown lol i got to the part: sin^2 (x) = -1/2 4. 4sin^3 (x) + 2sin^2 (x) - 2sin^2 (x) = 1 Solve the equation on the interval 0 less than or equals theta less than 2 pi 0 Sin(2x)=2sin(x)cos(x) find the exact value in radians with 2sin^2(x)+sin^2(x)=0, i really do not get this question or even how to do this ive been trying to figure it out but cant Prove the following identity: 1/tanx + tanx = 1/sinxcosx I can't seem to prove it. This is my work, I must've made a mistake somewhere: Converted 1/tanx: 1/sinx/cosx + sinx/cosx = 1/sinxcosx Simplified 1/sinx/cosx: cosx/sinx + sinx/cosx = 1/sinxcosx Found Hello...i had a question. I would really appreciate it if someone could help me...tnx in advance! K my question is Find the T(t) , N(t), B(t) for r(t)=; t=pie/4 ...which are normal vectors, binomial vector and i am not sure about T(T)...anyhow...i have the Which functions of x and y in terms of time t can be derived from this rectangular equation? (x^2/4)+y^2=1 A. x = 2sin t, y = -cos t B. x = sin t, y = 5cos t C. x = sin t, y = 2cos t D. x = -sin t, y = 2cos t E. x = 2sin t, y = 5cos t Which expression is equivalent to sin(3x) + sin x? A. 2cos(2x)sin x B. 2sin(2x)sin x C. -2sin(2x)cos x D. 2sin(2x)cos x E. -2cos(2x)sin x What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x) - 1 cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1 cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) If f(x)= (4cos^4x+2cos^2x+2sin^2x-x^9-1/2cos4x)^(1/9). Find f(f(2)) how would you solve for x: (3)^.5sin(2x)-2cos(2x)+2sin^2(x)=1 using trig identites. please help The graph r(t)= . For which values of t is the curvature largest? y = 2[cos1/2(f1 - f2)x][cos1/2(f1 + f2)x] let f1 = 16 let f2 = 12 therefore y = 2[cos1/2(16 - 12)x][cos1/2(16 + 12)x] y = 2cos(2x)cos(14x) derive y = 2cos(2x)cos(14x) y' = 2[-2sin(2x)][-14sin(14x)] is all of this correct? Additionally, I am unsure how you i have some problems doing trig the first one is: Show that cos(x/2) sin(3x/2) = ?(sinx + sin2x) i know that you are supposed to substitute all those trig function things in it but i kind of forgot how to the only that i can see substituting in is the Hello guys. I am just wondering if tan^2 A (tan squared A) = sin^2 A /cos^2A And 2tan A = 2sin A / 2cos A This has been bothering me when I have to simplify trig. Thanks in advance. Why does cos^2(x)= 1/2cos(2x)+1/2? I am trying to integrate, but the answer key says to first rewrite the expression like the above. I don't get how to change cos^2(x) into that. Explain? Determine the solution of the equetion: 2cos x-1=0? 2cos x=0? and 2cos x+1=0? 6.] Replace the integral in exercise 5 (int. (1/ 1 ? t) dt a = 0, b = 1/2with ?1/(1+t) dt with a = 0, b = 1, and repeat the four steps. a. integrate using a graphing utility b. integrate exactly c. integrate by replacing the integrand with a Taylor this is for proving identies and its fustrating i can't solve this one question! lol x=feta (btw the first part is supposed to be divided by the bottom) 1 + 2sinxcosx + sinxcosx sinx + cosx (1 + 2sinxcosx / sinx + cosx) + sinxcosx integrate from 0 to pi/4 (sec^2x)/((1+7tanx)^2)^1/3 integrate form pi^2/36 to pi^2/4 (cos(x^1/2))/(xsin(x^1/2))^1/2 integrate from 0 to pi/3 (tanx)/(2secx)^1/2 find the maximum value of f(x)= 3cos(x)-2sin(x) I know you have to take the derivative but after the derivative I don't know how to solve it. f'(x)=-3sin(x)-2cos(x)=0 I don't know what to do after this please help. 1-2sin^2(theta/2)=2cos^2(theta/2)-1 I am having problems with proving identities that has a number and an exponent with it. Like: *2cosxcsc2x = cscx *sin^2 2x = 4sin^2 x cos^2 x *cot2(theta)=csc(theta)-2sin(theta)/2cos(theta) Justify, with a written explanation or a mathematical reasoning and with a sketch of at least two different cases, the following properties of integrals: a) If f(x) is less than or equal to g(x) for a Need help solving problems using the Trigonometric Identities: 8cos^4 thet=cos4theta+4cos2theta+3 I am a bit unsure how to solve it. I am aware that the first step I must take is to simplify the problem and then use the Product to Sum formula Nasty one! 2sin(x)cos(x)+cos(x)=0 I'm looking for exact value solutions in [0, 3] So I need to find general solutions to solve the equation. But do I eliminate cos(x), like this... 2sin(x)cos(x)+cos(x)=0 2sin(x)cos(x)= -cos(x) 2sin(x) = -1 sin(x) = -1/2 at 4pi/3 Please can anyone help with the following problems - thanks. 1) Integrate X^4 e^x dx 2) Integrate Cos^5(x) dx 3) Integrate Cos^n(x) dx 4) Integrate e^(ax)Sinbx dx 5) Integrate 5xCos3x dx The standard way to solve most of these integrals is using partial dy/dx = 2y^2 Integrating...y=2/3 y^3 + C put 1,-1 into the equation, and solve for C. Then find the y for x=2 if y= a^u see exponential functions. dy/dx=2y^2 and if y=-1 when x=1, then when x=2, y=? how do i get x y= 2sin^2 x y=1- sinx find values of x inthe interval 0 solve each equation for 0=/ Integrate e^(2x)*((2x-1)/(4(x)^2))^2 My thoughts on this question : I simplfied the terms after the "*" to get three separate integrals : integrate (e^(2x)) + ((e^(2x))/4(x^2)) + ((e^(2x))/x) The answer for the first integral is obvious and then I was able Solve for x: 1. 2sin(2x)+cos(x)=0 2. cos(2x)=-2sin(x) 3. tan(x)=2sin(x) Find minimum value of 2cos x + 2cos y Find minimum value of 2cos x + 2cos y Find minimum value of 2cos x + 2cos y Sin(2x)=2sin(x)cos(x) find the exact value in radians with 2sin^2(x)+sin^2(x)=0, i really do not get this question or even how to do this ive been trying to figure it out but cant Page 2 am I right? 1. Simplify radical expression sqrt 50 5 sqrt ^2*** 2 sqrt ^5 5 sqrt ^10 5 2. Simplify the radical expression sqrt 56x^2 28x 2x sqrt 14*** 2x sqrt 7 sqrt 14x2 3. Simplify the radical expression. sqrt 490y^5w^6 2 sqrt 135y^2 2y sqrt 135*** 27 find the volume bounded by the parabolic cylinder z=4-x^2 and the planes x=0, y=0, y=6 and z=0 I just want some help figuring out the limits of this question : So for z, we have to integrate from z=0 to z=4-x^2 (am i right?) W.r.t. y, we should integrate Consider the area between the graphs x+y=16 and x+4= (y^2). This area can be computed in two different ways using integrals. First of all it can be computed as a sum of two integrals integrate from a to b of f(x)dx + integrate from b to c of g(x)dx What is integrate the following (3y-7x-3)dx+(7y-3x-7)dy=0 integrate du/3u and integrate x/x^2 dx it might be simple...i just need a head start...tnx A clothing store is having a sale on t-shirts: Buy 4 shirts at full price, then every subsequent shirt is half off. The total amount a customer spends on x shirts can be modeled by the piecewise function below: T(x) = 28x for 0 x 4 14x + 56 for x > Integrate: f (x)/(2x + 1) dx let f represent integrate sign let u = x, du = dx => dx = du = f (u)/(2u + 1) du = (2u + 1)^(-1) du = (1/2)u^2 (ln|2u + 1|) + c = (1/2)x^2 (ln|2x + 1|) + c ...what did I do wrong? The correct answer is (1/2)x - (1/4)ln|2x + 1| integrate (x^2-x+12)/(x^3+2x)dx Sketch the regions enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. integrate with either respect to x or y, then find area S of the region given that Integrate sin6xcos5xdx? Consider the polynomial p(x)=7x3+4x2-112x-64. Part A: What is the complete factorization of p(x)=7x3+4x2-112x-64 over the integers? Part B: What methods are used to factor p(x)=7x3+4x2-112x-64? Select one answer for Part A and select all Integrate from 1 to 5 of (3x-5)^5 dx = Integrate from a to b of f(u) du where (I have solved this part) u = 3x-5 du = 3 a = 0 b = 12 The original value of the integral is 165888 via calculator here's my last question, and it has to be in terms of u: f(u) = Sketch the regions enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. integrate with either respect to x or y, then find area S of the region given that dy/dx = 4ye^(5x) a) Separate the differential equation, then integrate both sides. b) Write the general solution as a function y(x). For the second part, I got y(x)=e^((5e^(5x))/(5)) + C but I don't understand how to separate differential equations and/or Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=5x , y=3 and 2y+1x=6 It is easier to integrate with respect to the variable Area = Help!!!! Practice problem: A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression: a= (2.8 m/s^3)t + (3.9 m/s^2) How far does the rocket move in the time interval t=0 to t=0.81 s? This is just a Integrate 2(x^2)e^g(x) where g(x)=4^(x^3) Integrate: dx/(2x^2 + 4x + 7) integrate x^2/(x^4+a^4) dx Integrate : (2x^2 ln8)dx integrate (e^t + e^-t)? Here is the graph. h t t p : / /goo.gl/PTc2I (spaces added at the beginning so it could be added as a website) 1. Let g be the function given by g(x)=integrate from -4 to x f(t)dt. For each of g(-1), g'(-1), and g''(-1), find the value of state that it Question : The length of time(in hundreds The length of time(in hundreds of hours) for the failure of a transistor is a random variable Y with distribution function ; F(Y) = { 0 : if y=0 } Find the probability that the transistor operates for at least 200 Question: Prove that [integrate {x*sin2x*sin[/2*cos x]} dx] /(2x-) } from (0-) = [ integrate {sin x*cos x*sin[/2*cos x} dx ] from (0-). My thoughts on the question: We know that integrate f(x) dx from (0-a) = integrate f(a-x) dx from (0-a) From 1)Solve by factoring:5x^2=4-19x answer=-4,1/5 2)Which quadratic equation has roots 7 and -2/3? answer=A A.2x^2-11x-21=0 B.3x^2-19x-14=0 C.3x^2+23x+14=0 D.2x^2+11x-21=0 3)To solve 4x^2-28x+49=25 by using the square root property,you would first rewrite the Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=4x^1/2,y=5,2y+1x=5 Do you really mean "2y+1x=5" ? It is not customary to use the coefficient 1 in front of a variable. Which of the following choices will most easily evaluate integral of [L(x)]/P(x) dx, where L(x) is a logarithmic function and P(x) is a polynomial? a) Integrate the polynomial function and differentiate the logarithmic function b) Integrate the logarithmic A chemical system within a sealed 1 L reaction vessel is described by the following reversible reaction equation: 2H2S(g) 2H(g) + S2(g) If the equilibrium constant is 0.000 004 200 at 1103 K find: (a) the reaction quotient intially (b) the order of dz =(-sinx + 2xy^2)dx +(2x^2 y)dy Integrate the differential to find the function z. Would I say that z was equal to two separate differentials and integrate the first part of the function with respect to x and the second part with respect to y? z = cosx should i use substitution?? if yes how should should i use it? plz i need some directions? k plz someone?...so far i used trig. substitution. i got a=8, so i used x=asin(?)so according to this substitution i got x=8sin(?) and dx=8cos(?) d?...then i Could someone explain how do we verify stokes theorem for the vector field F=zi + (2x+z)j + xk taken over the triangular surface S in the plane (x/1)+(y/2)+(z/3)=1 bounded by the planes x=0 y=0 and z=0. Take boundary of the above triangular surface as the How do you integrate 3(1)/(x?+1)? dx ? Thank you! integrate (e^t + e^-t)? integrate 2/((2x-7)^2) from x=4 to 6 how do you integrate x^2/(9+x^6) Integrate from [0, 1/2]: 1/(4x^2+1)^(3/2) dx integrate (x^2 + 2x + 5) / (x - 2) integrate (x-2) / (x+1)^2 +4 dx how do you integrate 3-e^1? Integrate:(K+1/k^2+4k+8)dx Please how do i integrate this = ?? (5 x + 2/x? ) ? Thanks integrate: (x^2 + 1)^k dx integrate (e^t + e^-t)? From 2 to 3, integrate ((x^3)+5)dx/x. HOW TO INTEGRATE : 2x(1+4x^2exp(-2x^2)) Thank you, Integrate xsinxcosx dx Integrate e^(2x)*((2x-1)/(4(x)^2))^2 My thoughts on this question : I simplfied the terms after the "*" to get three separate integrals : integrate (e^(2x)) + ((e^(2x))/4(x^2)) + ((e^(2x))/x) The answer for the first integral is obvious and then I was able Please help me integrate this equation using partial fractions: Integrate [(x^2+5)/(x^3-x^2+x+3)]dx. Thank you very much. How do you integrate [(x^2)(cos(2(x^3)))]? I tried to integrate by parts but I'm going in circles yet again... Find the greatest value of a,so that integrate [x*root{(a^2-x^2)/(a^2+x^2)} ] from 0-a 1) Integrate (e^x+x)^2(e^x+1) dx 2) Integrate xe^x2 dx Let u=(e^x+x) du=(e^x +1) dx I will be happy to critique your work or thinking. You are posting work for me to do, and I am not inclined to do that, it will not help you for me to do it. I think I gave what do you get when you: integrate 10sin^4xcosx dx when you integrate 9x^2e^6x^3 dx integrate (3+x)x^(1/2)dx=? please help I have no idea how to integrate this problem Integrate sqrt(x^2 + 1) dx over [0,2*pi] I can substitute u=arctan x to get: Integrate (sec u)^3 du over [0,arctan(2*pi)] From there, I'm stuck. (thanks Count Iblis for your last help) I need help with integrating these two problems. Im stuck. 1. integrate (sin^-1)dx/((1-x^2)^3/2) sin^-1 aka arcsin 2. integrate dx/((1-x^2)^3/2) by using 1/z Any and all help will be appreciated! Calculate the area bounded by the x-axis and the function f(x)= -(x-a)(x-b), where a integrate t*(t^2 1)^(1/3) dt over (0,3) I substitute u = t^2 - 1 du = 2t dt which leads to integrate (1/2) u^(1/3) du over(-1,8) = (3/8) * u^(4/3) over (-1/8) = 3/8 * [8^(4/3) - (-1)^(4/3)] I would guess that (-1)^(4/3) is +1, since the cube root of -1 Problems, once again. 1. Compute the average value of: f(x} = x/(x+3) over the interval [-a,a] 2. Find the area of the region bounded by the graph of: y = 2(x^2 + 1) X axis Y axis Line x = 1 On the first, integrate, then divide the integral by 2a. On Which of the following choices will most easily evaluate the integral of [L(x)]/[P(x)] dx, where L(x) is a logarithmic function and P(x) is a polynomial? a) Integrate the polynomial function and differentiate the logarithmic function. b) Integrate the integrate by parts integrate (4+x^2)^1/2 integrate by parts integrate (4+x^2)^1/2 I had to integrate [(x^2+1)/(x^2-x)]dx with partial fractions. My answer was 2ln abs(x-1) -ln abs(x)+C. But the answer on the answer sheet has an extra +x that I did not account for. Is that a typo or did I integrate incorrectly? Find the area bounded by the parabola y^2=4x and the line y=2x-4. Use vertical representative rectangles (integrate with respect to x) and horizontal representative rectangles (integrate with respect to y). the answer is 9 square units ... i just need to Find the area bounded by the parabola y^2=4x and the line y=2x-4. Use vertical representative rectangles (integrate with respect to x) and horizontal representative rectangles (integrate with respect to y). the answer is 9 square units ... i just need to I cannot for the life of me figure this out. Please help me. How do I integrate the function f(x) = 0.1 * e ^ (-0.2 * |x|) from neg. Infinity to pos. Infinity? I seem to only be able to get 0, but the answer is 1. I think it is the |x| that is throwing me I've been working on this hw problem for a while now, but I'm stuck in the integration process. I'm pretty sure I made an error, cause I can't seem to be able to integrate the right side of the equation. Q: (1/(x^(2)+1))y' + xy = 3 using the equation integrate -2/xln^4(x)...plz help me..give me an idea on how to start..plz The derivative of the ln(x) function is 1/x and this is multiplying the ln^4(x). You can thus write the integral as: -2 * 1/5 ln^5(x) + constant. is that the answer? wut about the x CASE STUDY : 2 The price P per unit at which a company can sell all that it produces is given by the function P(x) = 300 -- 4x. The cost function is c(x) = 500 + 28x where x is the number of units produced. Find x so that the profit is maximum. Question: 1. Solve using the Elimination Method. 2r - 7s = -57 7r %2B 2s = 39 What is the solution set ______? 2. Simplify -8[-90-(-60-14)]= ______ 3. Solve using the Substition Method. 5x %2B 7y = 41 x = 61 - 8y What is the solution _______? 4. On three consecutive Consider the function f(x)=-((x^2)/2)-9. In this problem you will calculate integrate from 0 to 3 of ((-x^2)/2)-9)dx by using the definition integrate from a to b of (f(x))dx= lim as n approaches infinity of sum_(i=1)^n of (f(x_i))(delta x) The summation Calc length of arc of y=ln(x) from x=1 to x=2 ---- So far: Definite Integral over x=(1,2) of sqrt(1 + 1/x) dx 1/x = tan^2 t x = 1/tan^2 t sqrt(1+1/x) = sqrt(1+tan^2 t) = sec t dx = -2 * tan^-3 t * sec^2 t dt Integrate over x=(1,2): sec^3 t / tan^3 t dt how do you integrate x^2/(9+x^6) how do you integrate ln(x^2+1)dx? how should I integrate 1/(x^2 +1)^5 how do you integrate dx/(x^2*(x^2+4)^1/2) I need to integrate ((e^-x)^2 - (-ex)^2)dx Can someone help? Integrate the following(6x^5-2x+1)dx Integrate ? (x^3)/?((x^2)+4) dx Integrate the following (x^5 - 2)dx how do you integrate dx/(x^2*(x^2+4)^1/2)? Integrate: x/(9+x^4)dx (ln(x))^2 dx, integrate. How do you integrate: 1/(x^2-1) dx ? integrate (5-4x^3+2x^6)/(x^6) Just not getting it. how do I integrate ((x^2)-(2/(1+x^4))dx How would you integrate x^2/x^2+9 ? how do I integrate x^4-3y^3+2y^2? thanks integrate (e^x-x)^2 dx Integrate (5x^2 + 3x - 2)/(x^3 + 2x^2) integrate x^2+4x+12/(x-2)(x^2+4) integrate (2/3x^5) integrate 2^(3x). 3^(2x) . 5^(x) dx help me integrate dx/(3x^3-5)^3 thanks Integrate:(3x^5-5)^-3 dx thanks How do you integrate e^(-t/r)? How do you integrate e^(-t/r) dt? integrate: (x^2 + 1)^k dx how do you integrate (3x+1)/(x-1)x^2 dx? CASE STUDY: 1 The bulbs manufactured by a company gave a mean life of 3000 hours with standard deviation of 400 hours. If a bulb is selected at random, what is the probability it will have a mean life less than 2000 hours? Question: 1) Calculate the Integrate e^(-t/r) WITH RESPECT TO dt?

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