Integration by Trigonometric Substitution

Key Questions

  • When integrating by trigonometric substitution, what are some useful identities to know?

    Useful Trigonometric Identities

    cos^2theta+sin^2theta=1

    1+tan^2theta=sec^2theta

    sin2theta=2sin theta cos theta

    cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta

    cos^2theta=1/2(1+cos2theta)

    sin^2theta=1/2(1-cos2theta)

    I hope that this was helpful.

    Wataru · · Oct 18 2014
  • How do you find the integral intdt/(sqrt(t^2-6t+13)) ?

    By completing the square,
    t^2-6t+13=(t^2-6t+9)+4=(t-3)^2+2^2

    So, we can rewrite the integral as
    int{dt}/{sqrt{(t-3)^2+2^2}}

    Let t-3=2tan theta.
    Rightarrow {dt}/{d theta}=2sec^2 theta Rightarrow dt=2sec^2 theta d theta

    by the above substitution,
    =int{2sec^2theta d theta}/{sqrt{(2tan theta)^2+2^2)} =int{sec^2theta}/{sqrt{tan^2theta+1}}d theta
    by the identity tan^2theta+1=sec^2theta,
    =int{sec^2theta}/{sqrt{sec^2theta}}d theta=int sec theta d theta=ln|sec theta + tan theta|+C_1
    by using
    tan theta={t-3}/2 and sec theta=sqrt{tan^2theta+1}={sqrt{t^2-6+13}}/2,
    we have
    =ln|{sqrt{t^2-6+13}}/2+{t-3}/2|+C_1
    =ln|{sqrt{t^2-6t+13}+t-3}/2|+C_1
    by log property,
    =ln|sqrt{t^2-6t+13}+t-3|-ln2+C_1
    by setting C=ln2+C_1,
    =ln|sqrt{t^2-6t+13}+t-3|+C

    Wataru · 2 · Sep 10 2014
  • How do we solve an integral using trigonometric substitution?

    In general trigonometric substitutions are useful to solve the integrals of algebraic functions containing radicals in the form sqrt(x^2+-a^2) or sqrt(a^2+-x^2). Consider the different cases:

    A. Let f(x) be a rational function of x and sqrt(x^2+a^2):

    int f(x)dx = int R(x, sqrt(x^2+a^2))dx

    Substitute: x = atant, dx = asec^2tdt with t in (-pi/2,pi/2) and use the trigonometric identity:

    1+tan^2t = sec^2t

    Considering that for t in (-pi/2,pi/2) the secant is positive:

    sqrt(x^2+a^2)= sqrt(a^2tan^2t+a^2)

    sqrt(x^2+a^2)=asqrt(tan^2t+1) = asect

    Then:

    int f(x)dx = int R(atant, asect)sec^2t dt

    B. Let f(x) be a rational function of x and sqrt(x^2-a^2):

    int f(x)dx = int R(x, sqrt(x^2-a^2))dx

    Restrict the function to x in (a,+oo) and substitute: x = asect, dx = asect tantdt with t in (0,pi/2) and use the trigonometric identity:

    sec^2t-1 = tan^2t

    Considering that for t in (0,pi/2) the tangent is positive:

    sqrt(x^2-a^2)= sqrt(a^2sec^2t-a^2)

    sqrt(x^2-a^2)=asqrt(sec^2t-1) = atant

    Then:

    int f(x)dx = int R(asect, atant)sect tant dt

    Normally you can see by differentiation that the solution that is found is valid also for x in (-oo, -a)

    C. Let f(x) be a rational function of x and sqrt(a^2-x^2):

    int f(x)dx = int R(x, sqrt(a^2-x^2))dx

    Substitute: x = a sint, dx = a cost with t in (-pi/2,pi/2) and use the trigonometric identity:

    1-sin^2t = cos^2t

    Considering that for t in (-pi/2,pi/2) the cosine is positive:

    sqrt(a^2-x^2)= sqrt(a^2-a^2sin^2t)

    sqrt(a^2-x^2)=a sqrt(1-sin^2t) = acost

    Then:

    int f(x)dx = int R(asint, acost)cost dt

    Andrea S. · 1 · May 25 2018

Questions

Techniques of Integration