Methods of integration

We have the following methods of integration:

 (i) Integration by substitution or change of independent of independent variable.

 (ii) Integration by parts.

 (iii) Integration of rotational algebraic function by using partial fractions.

Integration by substitution: If φ (x) is a continuous differentiable function, then to evaluate integrals of the form ∫f (φ(x)) φ’ (x) dx,

We substitute φ (x) = t and φ’ (x) dx = dt

We substitution reduces the integral ∫f(φ(x)) φ’ (x) dx

This substitution reduces the integral ∫f(φ(x)) φ’ (x) dx to the form ∫f (t) dt

∫f (φ(x)) φ’ (x) dx = ∫f (φ(x)) d (φ (x)) [∵ dφ (x) = φ’ (x) dx]

= ∫f (t) dt, where t = φ(x)

Example: e^x(x + 1) cos² (x.e^x) dx

Solution: we observe that e^x (x + 1) occurs in the derivative x.e^x. So let us substitute x.e^x = t

⇒ d (x.e^x).dx = dt

⇒ (x + 1) e^x dx = dt

⇒ 

∴ ∫ e^x(x + 1) cos² (x.e^x) dx

= ∫cos²t dt

= ½ ∫ (1 + cos2t) dt

Algorithm to evaluate integrals of the form: ∫sin^mx cosⁿ x dx, ∫sin^m x dx and ∫cosⁿ x dx, where m, n ϵ N

Step I: find m and n.

Step II: if m is odd i.e., power or index of sin x is odd, put cosx = t and reduce the integral in terms of t.

If n is odd i.e., power of cosx is odd, then put sinx=t, and reduce the integral in terms of t.

If m and n both are odd, then either of the above substitutions can be used.

Step III: evaluate the integral obtained in step II and replace t by its value.

Example: ∫x² sin³ x³ cos⁵³ dx

Solution: let I = ∫x² sin³ x³ cos⁵³ dx

Here, indices of both sine and cosine are odd. So, we may substitute sinx³ = t or cos³ = t.

Let sinx³ = t then,

d (sinx³) = dt

⇒ cosx³. 3x² dx = dt

⇒ 

∴ 

= ⅓ ∫t³ cos⁴ x³ dt

= ⅓ ∫t³ (1 - sin² x³)² dt

= ⅓ ∫t³ (1 - t²)² dt

= ⅓ ∫(t³ - 2t⁵ + t⁷) dt