Integration by Trigonometric Substitution

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Integrals by Trig Substitution Examples

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Key Questions

  • 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#

  • 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.

  • 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#

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