C. INTEGRATION BY SUBSTITUTION
Let g be a function whose range is an interval l, and let f be a function that is continuous on l. If g is differentiable on its domain and F is an antiderivative of f on l, then f(g(x))g'(x) dx = F(g(x)) + C.
If u = g(x), then du = g'(x) and f(u) du = F(u) + C .
GUIDELINES FOR MAKING A CHANGE OF VARIABLE
1. Choose a substitution u = g(x). Usually, it is best to choose the inner part of a composite function, such as a quantity raised to a power.
2. Compute du = g'(x) dx.
3. Rewrite the integral in terms of the variable u.
4. Evaluate the resulting integral in terms of u.
5. Replace u by g(x) to obtain an antiderivative in terms of x.
THE GENERAL POSER RULE FOR INTEGRATION
If g is a differentiable function of x, then
Some irrational functions can be changed into rational functions by means of appropriate substitutions.
In particular, when an integrand contains an expression of the form then the substitution u = may be effective.
SOME STANDARD SUBSTITUTIONS
Ex.8 Evaluate (x2 +1)2 (2x) dx .
Sol. Letting g(x) = x2 + 1, we obtain g'(x) = 2x and f(g(x)) = [g(x)]2.
From this, we can recognize that the integrand and follows the f(g(x)) g'(x) pattern. Thus, we can write
Let u = x4 + 2 ⇒ du = 4x3 dx
Let u = x3 – 2. Then du = 3x2 dx. so by substitution :
Sol. Let u = . Then u2 = x + 4, so x = u2 –4 and dx = 2u du.
Sol. Rewrite the integrand as follows :
= – ln (e-x + 1) + c (∴ e-x + 1 > 0)
Ex.14 Evaluate sec x dx
Sol. Multiply the integrand sec x by sec x + tan x and divide by the same quantity :
Ex.15 Evaluate cos x (4 - sin2 x) dx
Sol. Put sin x = t so that cos x dx = dt. Then the given integral =
Ex.20 Integrate cos5x.
[put sin x = t ⇒ cos x dx = dt]
Ex.22 Integrate 1/(sin3 x cos5x).
Sol. Here the integrand is sin–3 x cos–5x. It is of type sinm x cosn x,where m + n = –3 –5 = –8 i.e., –ve even integer
Now put tan x = t so that sec2 x dx = dt
Sol. Here the integrand is of the type cosm x sinnx. We have m = –3/2, n = – 5/2, m + n = – 4 i.e., and even negative integer.
,putting tan x = t and sec2x dx = dt
Put x – β = y ⇒ dx = dy
Now put sinθ + cosθ tan y = z2 ⇒ cosq sec2 y dy = 2z dz
Ex.25 Evaluate dx