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 $