Calculus Please Check my answer

find d/dx (integral from 2 to x^4) tan(x^2) dx
tan(x^4)^2*4x^3 MY ANSWER

asked by Ke$ha
  1. I assume you meant

    d/dx ∫[2,x^4] tan(t^2) dt
    = tan((x^4)^2) * 4x^3
    = 4x^3 tan(x^8)

    posted by Steve
  2. yes! thank you!

    posted by Ke$ha

Respond to this Question

First Name

Your Response

Similar Questions

  1. Trigonometry

    Find the exact value of tan(a-b) sin a = 4/5, -3pi/2<a<-pi; tan b = -sqrt2, pi/2<b<pi identity used is: tan(a-b)=(tan a-tan b)/1+tan a tan b simplify answer using radicals. (a is alpha, b is beta)
  2. Tan^2 Integrals

    I'm having a hard time understanding how to do Integrals involving tan^2. I have two problems: 1. Find the integral of (tan^2 y +1)dy 2. Find the integral of (7tan^2 u +15)du 1. My approach to it is to replace the tan^2 y portion
  3. Integration

    Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct? thanks. = ¡ì sec x d tan x = sec x tan x - ¡ì tan x d sec x = sec x tan x - ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x dx - ¡ì
  4. Calc

    intergral of (x^3 -4x + 3)/(2x) dx would this be ln abs(x^3-4x+3) + C? I don't really understand how to solve this problem. d/dx (tan (x^2)) sec^2(x^2)(2x) would this be the correct answer to find the derivative of tan (x^2)? no,
  5. Maths- complex numbers

    Find tan(3 theta) in terms of tan theta Use the formula tan (a + b) = (tan a + tan b)/[1 - tan a tan b) in two steps. First, let a = b = theta and get a formula for tan (2 theta). tan (2 theta) = 2 tan theta/[(1 - tan theta)^2]
  6. 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 -
  7. 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
  8. trig

    h t t p : / / i m g 4 0 . i m a g e s h a c k . u s / c o n t e n t . p h p ? p a g e = d o n e& l = i m g 4 0 / 1 4 8 / a s d a s d y e . j p g & v i a = m u p l o a d (def of tan theta = a^-1 o)a = o opposite = adjacent tan
  9. Math-Trigonometry

    Show that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C. I tried drawing perpendiculars and stuff but it doesn't seem to work? For me, the trig identities don't seem to plug in
  10. integral confusion

    integral of Sec[2x]Tan[2x] i know u is sec 2x du=2sec2xtan2x dx what would i have to multiply with du so it would equal tan 2x dx? if my question is confusing, then here's another example of what i'm talking about: integral of
  11. Calculus

    Calculate the following integral: ∫ sec^4 (3x)/ tan^3 (3x) dx For this one, can I bring up the tan to tan^-3?

More Similar Questions