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Integration by substitution : Integrals of certain functions can not be obtained directly if they are not in standard forms but they may be reduced to standard forms by proper substitution. The method of evaluating an integral by reducing it to standard form by a proper substitution is calledIf $\phi(x)$ is a continuously differentiable function, then to evaluate integrals of the form

$\int_{}^{} f(\phi(x)) \phi'(x) dx, $

We substitute $\phi(x)$ = u and $\phi'(x) dx = du$

This substitution reduces the above integral to

$\int_{}^{} f(u)du$

After evaluating this integral we substitute back the value of u.

2) $\int_{}^{}(ax +b)^{n}dx = \frac{(ax+b)^{(n+1)}}{a(n+1)} $ + C ,n $\neq$ -1

3) $\int_{}^{}sin(ax +b)dx = \frac{-1}{a} cos(ax+b) + c$

**Example 1 :** Integrate $(2x - 8)^{4}$ with respect to x.

**Solution : ** $\int_{}^{} (2x - 8)^{4}dx$

As the given function is not in the standard form so we can not find the integration of it directly. So we will use U-substitution method.

Let (2x - 8) = u

2dx = du

dx = $\frac{1}{2}du$

$\int_{}^{} (2x - 8)^{4}dx = \int_{}^{} (u)^{4}.\frac{1}{2}du$

= $\frac{1}{2}\int_{}^{} (u)^{4}du $

= $\frac{1}{2}.\frac{u^{4+1}}{4+1} $ (Applying power rule of integration)

= $\frac{1}{2}.\frac{u^{5}}{5} $

Now again replacing u as (2x-8)

$\int_{}^{} (2x - 8)^{4}dx = \frac{1}{2}.\frac{(2x-8)^{5}}{5} $

**Example 2 :** Integrate $sec^{2}(7 - 4x)$ with respect to x.

**Solution : ** $\int_{}^{} sec^{2}(7 - 4x)dx$

As the given function is not in the standard form so we can not find the integration of it directly. So we will use U-substitution method.

Let (7- 4x) = u

-4dx = du

dx = $\frac{-1}{4}du$

$\int_{}^{} sec^{2}(7 - 4x)dx = \int_{}^{} sec^{2}(u).\frac{-1}{4}du$

= $\frac{-1}{4}\int_{}^{} sec^{2}(u)du $

= $\frac{-1}{4}.tan(u) $ (Applying trigonometric integration rule)

Now again replacing u as (7 - 4x)

$\int_{}^{} sec^{2}(7 - 4x)dx = \frac{-1}{4}.tan(7-4x) + c $

**Example 3 :** Integrate $ sin^{2}(3x)cos(3x)$ with respect to x.

**Solution : ** $\int_{}^{} sin^{2}(3x)cos(3x)dx$

As the given function is not in the standard form so we can not find the integration of it directly. So we will use U-substitution method.

Let sin(3x) = u

3cos(3x) dx = du

cos(3x) dx = $\frac{1}{3}du$

$\int_{}^{} sin^{2}(3x)cos(3x)dx = \int_{}^{} u^{2}.\frac{1}{3}du$

= $\frac{1}{3}\int_{}^{} u^{2}du $

= $\frac{1}{3}.\frac{u^{2+1}}{2+1} $ (Applying power rule of integration)

=$\frac{1}{3}.\frac{u^{3}}{3} $

=$\frac{u^{3}}{9} $

Now again replacing u as sin(3x)

$\int_{}^{} sin^{2}(3x)cos(3x)dx = \frac{1}{9}.sin^{3}(3x) + c $

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