## Substitution Method in integration

In this page substitution method in integration we are going see where we need to use this method in integration. In this method we need to change the function which is defined one variable to another variable. If we have limits then we have to change that too.

Now let us see some example problems to understand this topic.

Example 1:

Integrate Cos x/(1 + sin x) with respect to x

Solution:

= ∫ [Cos x/(1 + sin x)] dx

Let U = 1 + sin x

Differentiate with respect to x on both sides

du = (0 + cos x) dx

du = cos x dx

∫ [Cos x/(1 + sin x)] dx =  ∫ [Cos x dx/(1 + sin x)]

=  ∫ du/u

=  ∫ (1/u) du

=  log u + C

=  log (1 + sin x) + C

Example 2:

Integrate (6x + 5)/√(3 x² + 5x + 6) with respect to x

Solution:

= ∫ [(6x + 5)/√(3 x² + 5x + 6)] dx

Let U =3 x² + 5x + 6

Differentiate with respect to x on both sides

du = (6x + 5 (1) + 0) dx

du = (6x + 5) dx

= ∫ (6x + 5) dx/√(3 x² + 5x + 6)

= ∫ du/√u

= ∫ u^(-1/2) du

=  u^[(-1/2) + 1]/[(-1/2) + 1] + C

=  u^[(-1 + 2)/2]/[(-1 + 2)/2] + C

=  u^(1/2)/(1/2) + C

=  (2/1) u^(1/2) + C

=  2 u^(1/2) + C

=  2 (3 x² + 5x + 6)^(1/2) + C

=  2 √(3 x² + 5x + 6) + C

Example 3:

Integrate Cosec x  with respect to x

Solution:

=  ∫ Cosec x dx

To solve this problem we have to multiply and divide by (cosec x - cot x)

=  ∫ Cosec x (cosec x - cot x)/(cosec x - cot x) dx

Let U=(cosec x - cot x)

Differentiate with respect to x on both sides

du = - Cosec x cot x - (- cosec² x)dx

du = - Cosec x cot x + cosec² x)dx

du =   Cosec²x- Cosec x cot x dx

=  ∫ (Cosec²x - Cosec x cot x) dx /(cosec x - cot x)

=  ∫ du/u

=  ∫ (1/u) du

=  log u + C

=  log (Cosec x - cot x) + C substitution method in integration  