Integration by Substitution & Trigonometric Identities

1. 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  

Rationalizing Substitutions

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

2. Integration Using Trigonometric Identities

In the integration of a function, if the integrand involves any kind of trigonometric function, then we use trigonometric identities to simplify the function that can be easily integrated.

Few of the trigonometric identities are as follows:

All these identities simplify integrand, that can be easily found out.

Solved Examples:

Ex.1 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

Ex.2 Evaluate  
Sol. 

Ex.3 Evaluate 
Sol.

Let u = x⁴ + 2   ⇒   du = 4x³ dx

Ex.4 Evaluate 

Sol.

Let u = x³ - 2. Then du = 3x² dx. so by substitution :

Ex.5 Evaluate  

Sol. Let u  =   . Then u² = x + 4, so x = u² -4 and dx = 2u du.

Therefore  

Ex.6 Evaluate  

Sol. Rewrite the integrand as follows :

 = - ln (e^-x + 1) + c   (∴  e^-x + 1 > 0)

Ex.7 Evaluate  sec x dx

Sol. Multiply the integrand sec x by sec x + tan x and divide by the same quantity :

Ex.8 Evaluate cos x (4 - sin2 x) dx
Sol. Put sin x = t so that cos x dx = dt. Then the given integral =

Ex.9 Integrate 

(i) 
(ii) 

Sol.

Ex.10 Integrate 

(i) 
(ii) 

Sol.

Ex.11 Integrate 

(i) 
(ii) 

Sol.

Ex.12 Integrate 

Sol.

Ex.13 Integrate cos⁵x.

Sol.

 [put sin x = t ⇒ cos x dx = dt] 

Ex.14 Evaluate 

Sol.

Ex.15 Integrate 1/(sin³ x cos⁵x).

Sol. Here the integrand is sin^-3 x cos^-5x. It is of type sin^m 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 

Ex.16 Integrate 

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

Ex.17 Evaluate  

Sol.

Put x - β = y  ⇒  dx = dy

Given integral 

Now put sinθ + cosθ tan y = z² ⇒  cosq sec² y dy = 2z dz

Ex.18 Evaluate dx

Sol.