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If $\phi(x)$ 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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