Case 1: Suppose we have an exponential function clubbed as ∫ex(f(x)+f′(x))dx. In this case, the integral is exf(x)+C.
Find the indefinite integral
∫ex(sin(x)+cos(x))dx,
using C as the constant of integration.
We have the integral in the form of ∫ex(f(x)+f′(x))dx, where f(x)=sin(x). So our integral is
exsin(x)+C. □
Case 2: Suppose we have an integration of the form I=∫eaxcos(bx+c). Its integral is
I=a2+b2eax(acos(bx+c)+bsin(bx+c)).
We'll integrate the above using integration by parts as follows:
I=∫eaxcos(bx+c) dx=cos(bx+c)aeax+ab∫eaxsin(bx+c) dx=cos(bx+c)aeax+ab(aeaxsin(bx+c)−ab∫eaxcos(bx+c)) dx=a2eax(acos(bx+c)+bsin(bx+c))−a2b2I.
So
I(1+a2b2)⇒I=a2eax(acos(bx+c)+bsin(bx+c))=a2+b2eax(acos(bx+c)+bsin(bx+c)). □
Note: The above example is also applicable for the form I=∫eaxsin(bx+c).
Find the indefinite integral
∫e2xcos(5x+3)dx,
using C as the constant of integration.
By the above result, we obtain our answer as
29e2x(2cos(5x+3)+5sin(5x+3))+C. □
Case 3: If the integration is of the form ∫pex+qe−xaex+be−xdx, express (NUM)=α(DEN)+βdxd(DEN), where NUM=(the numerator of the integrand) and DEN=(the denominator of the integrand), and then integrate as usual.
Find the indefinite integral
∫ex−5e−x2ex+3e−xdx,
using C as the constant of integration.
We can write 2ex+3e−x=α(ex−5e−x)+β(ex+5e−x). Comparing the coefficients of ex and e−x, we obtain α+β=2 and α−β=−53, which implies α=107,β=1013. So we have
∫ex−5e−x2ex+3e−xdx=α∫dx+β∫ex−5e−xex+5e−xdx.(1)
Let ex−5e−x=t, then (ex+5e−x)dx=dt, which gives
(1)=107∫dx+1013∫tdt=107x+1013ln∣t∣+C=107x+1013ln∣∣ex−5e−x∣∣+C,
where C is the constant of integration. □