# 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: ex(x + 1) cos² (x.ex) dx

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

⇒ d (x.ex).dx = dt

⇒ (x + 1) ex dx = dt

⇒ $$dx=\frac{dt}{\left( x+1 \right){{e}^{x}}}$$

∴ ∫ ex(x + 1) cos² (x.ex) dx

= ∫cos²t dt

= ½ ∫ (1 + cos2t) dt

= $$\frac{1}{2}\left[ t+\frac{\sin 2t}{2} \right]+C$$

= $$\frac{1}{2}t+\frac{1}{4}\sin 2t+C$$

= $$\frac{1}{2}x{{e}^{x}}+\frac{1}{4}\sin 2\left( x{{e}^{x}} \right)+C$$

Algorithm to evaluate integrals of the form: ∫sinmx cosn x dx, ∫sinm 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

⇒ $$dx=\frac{dt}{3{{x}^{2}}\cos {{x}^{3}}}$$

∴ $$I=\int {{x}^{2}}{{t}^{3}}{{\cos }^{5}}{{x}^{3}}\frac{dt}{3{{x}^{2}}\cos {{x}^{3}}}$$

= ⅓ ∫t³ cos⁴ x³ dt

= ⅓ ∫t³ (1 – sin² x³)² dt

= ⅓ ∫t³ (1 – t²)² dt

= ⅓ ∫(t³ – 2t⁵ + t⁷) dt

= $$\frac{1}{3}\left[ \frac{{{t}^{4}}}{4}-\frac{2{{t}^{6}}}{6}+\frac{{{t}^{8}}}{8} \right]+C$$

= $$\frac{1}{12}{{\sin }^{4}}{{x}^{3}}-\frac{1}{9}{{\sin }^{6}}{{x}^{3}}+\frac{1}{24}{{\sin }^{8}}{{x}^{3}}+C$$