Integral problem, need help?
∫2ytanֿ¹(6y) dy = 2tanֿ¹(6y) ∫y dy - 2 ∫[d/dy tanֿ¹(6y) ∫y dy] dy = y^2tanֿ¹(6y) - ∫6y^2 / (1+36y^2) dy = y^2tanֿ¹(6y) - (1/6) ∫(1+36y^2-1)/(1+36y^2) dy = y^2tanֿ¹(6y) - (1/6) [∫dy - ∫1/(1+36y^2) dy] = y^2tanֿ¹(6y) - (1/6) [ y - (1/6)tanֿ¹(6y) ] + c = y^2tanֿ¹(6y) - (1/6)y + (1/36) tanֿ¹(6y) + c. ∫xsinֿ¹(4x) dx Let sinֿ¹(4x) = t => 4x = sint and 4dx = cost dt Also x=0 => t=0 and x=1/4 => t=π/2 => Integral = ∫(1/4) sint * t * (1/4) cost dt = (1/32) ∫t sin2t dt = (1/32) [ t∫sin2t dt + (1/2)∫cos2t dt] = - (1/64)tcos2t + (1/128)sin2t = - (1/64)(π/2)(-1) = π/128. Mohanrao has made an error in the step =(1/2) x^2 arcsin(4x) + 1/8 ∫ dx - 1/8 ∫ dx /√(1 - 16x^2) It should be = =(1/2) x^2 arcsin(4x) + 1/8 ∫√(1 - 16x^2)dx - 1/8 ∫ dx /√(1 - 16x^2)
Help with integral problems?
A stone is thrown vertically upward from the top of a house 60 feet above the ground with the initial velocity of 40 ft/sec. a) At what time will the stone reach its maximum height? b) What is its greatest height? c) How long will it take the stone to pass the top of the house on its way down? d) What is the velocity at that instant? e) How long will it take the stone to strike the ground? f) What is the velocity when it strikes the ground?
Help with an integral problem!?
(A) Let u = 8t, then du = 8dt => dt = (1/8) du This change of variable will yield 19/8 Similarly, use change of variable for (B) and (C)
Please help me with these integral problems!?
(1) and (2) test your memory of various trignometric derivatives. The derivative of csc u is -csc u cos u, and the derivative of cot u is -csc^2 u. So you can take the first integral by substituting u = 3x and then noticing that -csc u is an antiderivative of csc u cot u; the result is (-1/3) csc(3x) (plus a constant). Similarly you can take the second integral by substituting u = 7x and then noticing that -cot u is an antiderivative of csc^2(u); the result is (-1/7) cot(7x) (plus a constant). In (3) I would use the trigonometric identity tan^2 t + 1 = sec^2 t to rewrite the integral as the integral of sec^2(5x) - 1 dx. You can integrate the first term using a substitution u = 5x and then recalling that an antiderivative of sec^2 u du is tan u. The integral of the second term is just x In (4) I would rewrite it as tan^2(5x) * sec(5x) * sec(5x) tan(5x) = (sec^2(5x) - 1) * sec(5x) * sec(5x) tan(5x) Then substitute u = sec(5x) and note that du = 5 sec(5x) tan(5x) dx. So it turns into the integral of (u^2 - 1) u (1/5) du It may all seem rather counterintuitive but it becomes clearer with practice what is a good idea and what isn't. The key is to remember the derivatives of all the trig functions (even the weird ones like csc and cot) and identities like tan^2 t + 1 = sec^2 t (which comes from dividing both sides of sin^2 t + cos^2 t = 1 by cos^2 t)
I need help with a crazy integral problem?
1 ∫ (ax)² e^(-x) dx = -1 let's find the antiderivative: ∫ (ax)² e^(-x) dx = first of all, let's rewrite the square as: ∫ (a²x²) e^(-x) dx = let: (a²x²) = u → 2a²x dx = du e^(-x) dx = dv → - e^(-x) = v thus, integrating by parts, you get: ∫ u dv = v u - ∫ v du → ∫ (a²x²) e^(-x) dx = [- e^(-x)] (a²x²) - ∫ [- e^(-x)] 2a²x dx = - a²x² e^(-x) + ∫ 2a²x e^(-x) dx = a further by parts integration being needed, let again e^(-x) dx = dv → - e^(-x) = v 2a²x = u → 2a² dx = du thus, integrating the remaining integral, you get: - a²x² e^(-x) + { ∫ 2a²x e^(-x) dx} = - a²x² e^(-x) + {[- e^(-x)] (2a²x) - ∫ [- e^(-x)] 2a²dx} = - a²x² e^(-x) + {- 2a²x e^(-x) + ∫ 2a² e^(-x)] dx} = pullhe constants out: - a²x² e^(-x) - 2a²x e^(-x) + 2a² ∫ e^(-x) dx = - a²x² e^(-x) - 2a²x e^(-x) + 2a² [- e^(-x)] + C thus the antiderivative is: ∫ (ax)² e^(-x) dx = - a²x² e^(-x) - 2a²x e^(-x) - 2a² e^(-x) + C finally, evaluating the definite integral, you get: 1 ∫ (ax)² e^(-x) dx = {- a²(1)² e^[- (1)] - 2a²(1) e^[- (1)] - 2a² e^[- (1)]} - -1 {- a²(-1)² e^[- (-1)] - 2a²(-1) e^[- (-1)] - 2a² e^[- (-1)]} = - a²e^(-1) - 2a²e^(-1) - 2a²e^(-1) - [- a²(1) (e^1) + 2a²(e^1) - 2a² (e^1)] = - 5a²e^(-1) + a²e I hope it helps... Bye!
Can you help me with this integration problem? How do I find the dimensions of the triangles?
I'm not sure exactly what the problem is asking for (can't see it in the screenshot) but I'll try to help.If you are trying to find the area of the shaded region you can solve the equation of the ellipse for y and then integrate the positive root from -3 to 3. This will give you half the area of the ellipse. Multiply that number by to for the whole area.If you are trying to find volume, you can use the washer / shell method, which you might want to review elsewhere.Hope that helps:)
Help with calculus integral problem: cos(x^5) and sin(x^6)?
Is this what you're looking for? cos(x5)sin(x6) Find the derivative of the expression. cos(x5)sin(x6) Use the product rule to find the derivative of cos(x5)sin(x6). The product rule states that (fg)'=f'g+fg'. sin(x6)] The derivative of cos(x5) is cos(x5)=-5x4sin(x5) Substitute the derivative back into the product rule formula. sin(x6)] The derivative of sin(x6) is sin(x6)=6x5cos(x6) Substitute the derivative back into the product rule formula. cos(x5)sin(x6)=(-5x4sin(x5))(sin(x6))+... Simplify the derivative. cos(x5)sin(x6)=-5x4sin(x5)sin(x6)+6x5c... The derivative of cos(x5)sin(x6) is -5x4sin(x5)sin(x6)+6x5cos(x5)cos(x6). -5x4sin(x5)sin(x6)+6x5cos(x5)cos(x6)
Help with math integration problem plz?
You are correct with the first integration. The second goes like this: tanx=sinx/cosx pick your u. u=cosx, du=-sinx. You could choose it the other way, but then your integration becomes messier. so now you have - S du/u. As you know that is Ln( abs(u) ). So there is your answer. Most mathbooks have that in the back of the book as an integration to memorize, so you dont have to show all your work all the time. You should ask your teacher if you should memorize or show your work all the time. In a problem like this he probably wants you to show your work.
I need some help with this trigonometric integral problem?
I'm just learning myself, so this is just a beginner's attempt: INt (1 - sin^2 x)^2 cosx sin^2 x let u = sin x, du = cos x then integral becomes u(1 - u^2)^2 du expanding: INT (u - 2u^3 + u^5) integrating: u^2/2 - (u^4)/2 + (u^6)/6 + c substituting for u: = (1/2)sin^2(x) - (1/2)sin^4(x) + (1/6)sin^6(x) + c here's some info on the subject: http://www.sosmath.com/calculus/integrat...