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 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# -
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#