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}<br />
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}