# Examples of Integration by Substitution

Example: Evaluate the integral $\int {\frac{{{{\sin }^3}\left( {\ln \ln x} \right)\cos \left( {\ln \ln x} \right)}}{{x\ln x}}} dx$ with respect to $x$.

We have integral $I = \int {\frac{{{{\sin }^3}\left( {\ln \ln x} \right)\cos \left( {\ln \ln x} \right)}}{{x\ln x}}} dx$

$I = \int {{{\sin }^3}\left( {\ln \ln x} \right)\cos \left( {\ln \ln x} \right)} \frac{1}{{x\ln x}}dx$

Putting $t = \ln \ln x$ and differentiating $dt = \frac{1}{{x\ln x}}dx$

Now the above integral of the form
$I = \int {{{\sin }^3}t} \cos tdt$

We observe that the derivation of given function is in the given problem, so using the general power formula of integration
$\int {{{\left[ {f\left( t \right)} \right]}^n}f’\left( t \right)} dt = \frac{{{{\left[ {f\left( t \right)} \right]}^{n + 1}}}}{{n + 1}} + c$

Here $f\left( t \right) = \sin x$ implies that $f’\left( t \right) = \cos x$

$\begin{gathered} I = \frac{{{{\sin }^{3 + 1}}t}}{{3 + 1}} + c \\ \Rightarrow I = \frac{1}{4}{\sin ^4}t + c \\ \end{gathered}$

Now using the original substitution again $t = \ln \ln x$ in the result of the integration, we have
$I = \frac{1}{4}{\sin ^4}\left( {\ln \ln x} \right) + c$

Example: Integrate $\frac{{{e^{\sqrt {x + 1} }}}}{{\sqrt {x + 1} }}$ with respect to $x$.

Consider the function to be integrate $I = \int {\frac{{{e^{\sqrt {x + 1} }}}}{{\sqrt {x + 1} }}dx}$

$I = \int {{e^{\sqrt {x + 1} }}\frac{1}{{\sqrt {x + 1} }}dx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Putting $t = \sqrt {x + 1}$ and differentiating $dt = \frac{1}{{2\sqrt {x + 1} }}dx$ implies $2dt = \frac{1}{{\sqrt {x + 1} }}dx$

Now by using these values, equation (i) becomes
$\begin{gathered} I = \int {{e^t}2dt} \\ \Rightarrow I = 2\int {{e^t}dt} \\ \end{gathered}$

Using the formula of integration $\int {{e^x}dx = {e^x} + c}$, we have
$\begin{gathered} I = 2{e^t} + c \\ \Rightarrow I = 2{e^{\sqrt {x + 1} }} + c \\ \end{gathered}$