Calculus AP

posted by .

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

∫ tan^3 (2x) sec^5(2x) dx
=∫ tan^2(2x) sec^4(2x) tan*sec(2x) dx
=∫ (sec^2(2x)-1)sec^4 tan*sec(2x) dx
let u=sec x, du= 1/2 tan*sec(2x) dx
=1/2∫ (u^2(2x)-1) u^4 du
=1/2∫ (u^8(2x)-u^4) du
=1/2 sec^9/9-sec^5/5 +c

  • Calculus AP -

    you dropped some 2's here and there, and the final integral is

    1/2∫ (u^2-1) u^4 du
    = 1/2 ∫ u^6 - u^4 du
    = 1/2 (1/7 u^7 - 1/5 u^5)
    = 1/14 sec^7(2x) - 1/10 sec^5(2x) + C

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

    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 …
  4. 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 …
  5. 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)] …
  6. 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?
  7. Calculus - Question

    Am I allowed to do this? for the integral of ∫ sec^4 (3x)/ tan^3 (3x) dx I change it to ∫ sec^4 (3x) tan^-3 (3x) From here I use the rule for trigonometry functions.
  8. calculus II

    ∫ tan^2 x sec^3 x dx If the power of the secant n is odd, and the power of the tangent m is even, then the tangent is expressed as the secant using the identity 1 + tan^2 x = sec^2 x I thought that since tan is even and sec is …
  9. Calculus 2

    ∫ tan^2 (x) sec^4 (x) dx ∫ [tan^2 (t) + tan^4 (t)] dt ∫ [1-tan^2 (x)] / [sec^2 (x)] dx Trigonometric integral Please show steps so I can understand!
  10. calculus trigonometric substitution

    ∫ dx/ (x^2+9)^2 dx set x = 3tan u dx = 3 sec^2 u du I = 3 sec^2 u du / ( 9 tan^2 u + 9)^2 = 3 sec^2 u du / ( 81 ( tan^2 u + 1)^2 = sec^2 u du / ( 27 ( sec^2 u )^2 = du / ( 27 sec^2 u = 2 cos^2 u du / 54 = ( 1 + cos 2u) du / 54 …

More Similar Questions