Sign In | Subscribe
INSTRUCTORS Raffi Hovasapian John Zhu
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Calculus AB
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Related Books

Bookmark and Share
Lecture Comments (7)

0 answers

Post by Arjun Chaturvedi on April 25, 2014

Can't see the rest of video.

0 answers

Post by Akilah Miller on February 10, 2014

I do not believe, for example 3, that the question involves the inverse trig identities on the account that it does not match any of them. So, I used u-substitution instead and the result was 4/3(x^2)^3/4 + C.  Is this correct or am I missing something?

1 answer

Last reply by: Arjun Chaturvedi
Fri Apr 25, 2014 8:54 PM

Post by sergey smith on October 29, 2013

wouldn't #3 be 2arcsinh(x)+C

0 answers

Post by Latricia Glavin on June 11, 2013

uhm I believe the correct answer is 4arctanx + c to example 3

1 answer

Last reply by: Colby Trace
Thu Apr 4, 2013 6:02 PM

Post by Shehan Gunasekara on June 11, 2012

Example 3 was great ;P

Integrals of Inverse Trigonometric Functions

  • Memorize!
  • Manipulate expression to resemble that of the inverse trig formula to apply

Integrals of Inverse Trigonometric Functions

∫[1/(√{x2 − √2})] dx
  • ∫[1/(√{x2 − √2})] dx = ∫[1/(√{x2 − (2[1/4])2})] dx
∫[1/(√{x2 − √2})] dx = sin−1 [x/(2[1/4])] + c
∫[cosx/(sin2 x + 4)] dx
  • u = sinx
  • du = cosx  dx
  • ∫[cosx/(sin2 x + 4)] dx = ∫[1/(u2 + 4)] du
  • ∫[cosx/(sin2 x + 4)] dx = ∫[1/(22 + u2)] du
  • ∫[cosx/(sin2 x + 4)] dx = [1/2] tan−1 [u/2] + c
∫[cosx/(sin2 x + 4)] dx = [1/2] tan−1 ([sinx/2]) + c
∫[(ex)/(√{e4x − e2x})] dx
  • ∫[(ex)/(√{e4x − e2x})] dx = ∫[(ex)/(√{(e2x)(e2x − 1)})] dx
  • ∫[(ex)/(√{e4x − e2x})] dx = ∫[(ex)/(ex √{e2x − 1})] dx
  • ∫[(ex)/(√{e4x − e2x})] dx = ∫[(ex)/(ex √{(ex)2 − 12})] dx
  • u = ex
  • du = ex dx
  • ∫[(ex)/(√{e4x − e2x})] dx = ∫[1/(u √{u2 − 12})] du
  • ∫[(ex)/(√{e4x − e2x})] dx = [1/1] sec−1 [u/1] + c
∫[(ex)/(√{e4x − e2x})] dx = sec−1 ex + c
∫[1/(√{−x2 − 2x})] dx
  • ∫[1/(√{−x2 − 2x})] dx = ∫[1/(√{1 − (x2 + 2x + 1)})] dx
  • ∫[1/(√{−x2 − 2x})] dx = ∫[1/(√{1 − (x + 1)2})] dx
  • u = x + 1
  • du = dx
  • ∫[1/(√{−x2 − 2x})] dx = ∫[1/(√{1 − u2})] du
  • ∫[1/(√{−x2 − 2x})] dx = sin−1 u + c
∫[1/(√{−x2 − 2x})] dx = sin−1 (x + 1) + c
∫[(x2)/(9 + x6)] dx
  • ∫[(x2)/(9 + x6)] dx = ∫[(x2)/(9 + (x3)2)] dx
  • u = x3
  • du = 3x2 dx
  • ∫[(x2)/(9 + x6)] dx = [1/3] ∫[1/(32 + (x3)2)] 3x2 dx
  • ∫[(x2)/(9 + x6)] dx = [1/3] ∫[1/(32 + u2)] dx
  • ∫[(x2)/(9 + x6)] dx = [1/3] [1/3] tan−1 [u/3] + c
∫[(x2)/(9 + x6)] dx = [1/9] tan−1 [(x3)/3] + c
∫[1/(x2 + 2x + 3)] dx
  • ∫[1/(x2 + 2x + 3)] dx = ∫[1/(x2 + 2x + 1 + 2)] dx
  • ∫[1/(x2 + 2x + 3)] dx = ∫[1/((x + 1)2 + √22)]
  • u = x + 1
  • du = dx
  • ∫[1/(x2 + 2x + 3)] dx = ∫[1/(√22 + u2)] du
  • ∫[1/(x2 + 2x + 3)] dx = [1/(√2)] tan−1 [u/(√2)] + c
∫[1/(x2 + 2x + 3)] dx = [1/(√2)] tan−1 [(x + 1)/(√2)] + c
∫[1/(√{4x2 − (x lnx)2})] dx
  • ∫[1/(√{4x2 − (x lnx)2})] dx = ∫[1/(√{x2(22 − (lnx)2})] dx
  • ∫[1/(√{4x2 − (x lnx)2})] dx = ∫[1/x] [1/(√{22 − (lnx)2})] dx
  • u = lnx
  • du = [1/x] dx
  • ∫[1/(√{4x2 − (x lnx)2})] dx = ∫[1/(√{22 − u2})] du
  • ∫[1/(√{4x2 − (x lnx)2})] dx = sin−1 [u/2] + c
∫[1/(√{4x2 − (x lnx)2})] dx = sin−1 [lnx/2] + c
∫[sinx/(√{16 − cos2(x)})] dx
  • u = cosx
  • du = −sinx  dx
  • ∫[sinx/(√{16 − cos2(x)})] dx = − ∫[1/(√{16 − cos2(x)})] (−sinx) dx
  • ∫[sinx/(√{16 − cos2(x)})] dx = − ∫[1/(√{42 − u2})] du
  • ∫[sinx/(√{16 − cos2(x)})] dx = − sin−1 [u/2] + c
∫[sinx/(√{16 − cos2(x)})] dx = − sin−1 [cosx/2] + c
∫[1/(x ln(5x)√{(ln(5x))2 − 25})] dx
  • u = 5x
  • du = 5  dx
  • ∫[1/(x ln(5x)√{(ln(5x))2 − 25})] dx = [1/5] ∫[1/(x lnu√{(lnu)2 − 52})] du
  • ∫[1/(x ln(5x)√{(ln(5x))2 − 25})] dx = [1/5] ∫[5/(u lnu √{(lnu)2 − 52})] du
  • v = lnu
  • dv = [1/u] du
  • ∫[1/(x ln(5x)√{(ln(5x))2 − 25})] dx = ∫[1/(v √{v2 − 52})] dv
  • ∫[1/(x ln(5x)√{(ln(5x))2 − 25})] dx = [1/5] sec−1 [v/5] + c
∫[1/(x ln(5x)√{(ln(5x))2 − 25})] dx = [1/5] sec−1 [ln5x/5] + c
∫[(x3)/((x4 + 1)2 + 36)] dx
  • u = x4 + 1
  • du = 4x3 dx
  • ∫[(x3)/((x4 + 1)2 + 36)] dx = [1/4] ∫[1/((x4 + 1)2 + 36)] 3x3 dx
  • ∫[(x3)/((x4 + 1)2 + 36)] dx = [1/4] ∫[1/((u2 + 62)] dx
  • ∫[(x3)/((x4 + 1)2 + 36)] dx = [1/4] [1/6] tan−1 [u/6] + c
∫[(x3)/((x4 + 1)2 + 36)] dx = [1/24] tan−1 [(x4 + 1)/6] + c

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Integrals of Inverse Trigonometric Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Types of Integrals: Inverse Trig Functions 0:09
    • One Property
  • Example 1 1:19
  • Example 2 3:44
  • Example 3 4:53
  • Example 4 5:53