Integration by Trigonometric Substitution
Key Questions
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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.
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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 identitytan^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 andsec 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 settingC=ln2+C_1 ,
=ln|sqrt{t^2-6t+13}+t-3|+C -
In general trigonometric substitutions are useful to solve the integrals of algebraic functions containing radicals in the form
sqrt(x^2+-a^2) orsqrt(a^2+-x^2) . Consider the different cases:A. Let
f(x) be a rational function ofx andsqrt(x^2+a^2) :int f(x)dx = int R(x, sqrt(x^2+a^2))dx Substitute:
x = atant ,dx = asec^2tdt witht 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 ofx andsqrt(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 witht 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 ofx andsqrt(a^2-x^2) :int f(x)dx = int R(x, sqrt(a^2-x^2))dx Substitute:
x = a sint ,dx = a cost witht 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