AP Calculus BC

posted by .

Hi! Thank you very much for your help---
I'm not sure what the answer to this is; how do I solve?

Find antiderivative of
(1/(x^2))[sec(1/x)][tan(1/x)]dx

I did integration by parts and got to
(1/(x^2))[sec(1/x)] + 2*[antiderivative of (1/(x^3))(sec(1/x))dx]

  • AP Calculus BC -

    Integration by parts is the same as any other tool. It's just a tool. You can go around in circles with it... unless you know where you're going.

    For this particular problem, I propose to use another tool, substitution.

    Did you notice there is the factor (1/x²) at the beginning? What would ∫(1/x²)dx give? ∫-d(1/x).

    So the integral becomes:
    I=∫(1/(x^2))[sec(1/x)][tan(1/x)]dx
    =∫[sec(1/x)][tan(1/x)]d(1/x)
    =∫sec(y)tan(y)dy
    = ... +C

    Do remember, however, if and when you have to evaluate a definite integral, the limits have to correspond to the integration variable, which in this case is (1/x).

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Integration

    Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct?
  2. Calculus

    could anybody please explain how sec x tan x - ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x dx - ¡ì sec^3(x) dx What I don't understand about your question is what is ¡ì ?
  3. Calculus - Integration

    Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you! 1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx It says u = tan x to substitute So if I use u = tan x, then my du = secx^2 then I …
  4. calculus

    find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x …
  5. calculus

    Use integration by parts to evaluate the integral of x*sec^2(3x). My answer is ([x*tan(3x)]/3)-[ln(sec(3x))/9] but it's incorrect. u=x dv=sec^2(3x)dx du=dx v=(1/3)tan(3x) [xtan(3x)]/3 - integral of(1/3)tan(3x)dx - (1/3)[ln(sec(3x))/3] …
  6. Calculus 12th grade (double check my work please)

    2- given the curve is described by the equation r=3cos ¥è, find the angle that the tangent line makes with the radius vector when ¥è=120¨¬. A. 30¨¬ B. 45¨¬ C. 60¨¬ D. 90¨¬ not sure A or D 2.) which of the following represents …
  7. calculus (check my work please)

    Not sure if it is right, I have check with the answer in the book and a few integral calculators but they seem to get a different answer ∫ sec^3(x)tan^3(x) dx ∫ sec^3(x)tan(x)(sec^2(x)-1) dx ∫ tan(x)sec(x)[sec^4(x)-sec^2(x)] …
  8. calculus

    So I am suppose to evaulate this problem y=tan^4(2x) and I am confused. my friend did this : 3 tan ^4 (2x) d sec^ 2x (2x)= 6 tan ^4 (2x) d sec^2 (2x) She says it's right but what confuses me is she deriving the 4 and made it a three?
  9. Calculus AP

    I'm doing trigonometric integrals i wanted to know im doing step is my answer right?
  10. Calculus

    How do I find the critical values? y= 4/x + tan(πx/8) What I did is I simplified it to y= 4x^-1 + tan(πx/8) then I took the derivative y'= -4x^-2 + (π/8)(sec(πx/8))^2 Then I simplied it y'= -4/x^2 + (π/8)(sec(πx/8))^2

More Similar Questions