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
\[\begin{gathered} I = 2{e^t} + c \\ \Rightarrow I = 2{e^{\sqrt {x + 1} }} + c \\ \end{gathered} \]