4.2 INTEGRATION BY SUBSTITUTION

So far we have dealt with functions, either directly integrable using integration formula (or) integrable after decomposing the given functions into sums & differences.

\[But\ there\ are\ functions\ like\ \frac{sin(log x)}{x},\ \frac{2x + 3}{x^2 + 3x + 5}\]

which cannot be decomposed into sums (or) differences of simple functions. In these cases, using proper substitution, we shall reduce the given form into standard form, which can be integrated using basic integration formula.

When the integrand (the function to be integrated) is either in multiplication or in division form and if the derivative of one full or meaningful part of the function is equal to the other function then the integration can be evaluated using substitution method as given in the following examples.

\[1.\ \int \frac{sin(log\ x)}{x}\ dx= \int sin (log\ x) \frac{1}{x}\ dx\]
\[Here\ \frac{d}{dx} (log\ x) = \frac{1}{x}\]

The above integration can be evaluated by taking  u = log x.

\[2.\ \int \frac{2x + 3}{x^2 + 3x + 5}\ dx\]
\[since\ \frac{d}{dx} ( x^2 + 3x + 5)\ is\ 2x + 3\]

it can be integrated by  taking  u  =x2 + 3x + 5.

Integrals of the form

\[\int[f(x)]^n\ f^!(x)\ dx,\ n\ \neq\ -1,\]
\[\int\ \frac{f^!(x)}{f(x)}\ dx\]

and

\[\int[F(f(x))]\ f^!(x)\ dx\]

are all, more or less of the same type and the use of substitution u = f(x) will reduce the given function to simple standard form which can be integrated using integration formulae.

Example:

\[Evaluate\ :\ \int\ (5x + 2)^6\ dx\]

Soln:

\[put\ u = 5x + 2 \]
\[\frac{du}{dx}= 5\]
\[\frac{1}{5}\ du= dx\]
\[\int\ (5x + 2)^6\ dx= \frac{1}{5} \int\ u^6\ du\]
\[=\frac{1}{5} \frac{u^7}{7} + c\]
\[=\frac{(5x + 2)^7}{35} + c\]

Example:

\[Evaluate\ :\ \int\ (x^2 + x + 1)^5\ (2x + 1) dx\]

Soln:

\[put\ u =x^2 + x + 1\]
\[\frac{du}{dx}= \ 2x + 1\]
\[\int\ (x^2 + x + 1)^5\ (2x + 1) dx= \int\ u^5\ du\]
\[=\frac{u^6}{6} + c\]
\[=\frac{(x^2 + x + 1)^6}{6} + c\]

Example:

\[Evaluate\ :\ \int\ \frac{2x}{1 + x^2}\ dx\]

Soln:

\[put\ u\ =\ 1 + x^2\]
\[\frac{du}{dx}= \ 2x\]

du  = 2 dx

\[\int\ \frac{2x}{1 + x^2}\ dx= \int\ \frac{du}{u}\]

=  log u + c

\[= log(1 + x^2) + c\]
\[\int\ \frac{2x}{1 + x^2}\ dx= log(1 + x^2) + c\]

Example:

\[Evaluate\ :\ \int\ \frac{dx}{x\ log x}\]

Soln:

\[\int\ \frac{dx}{x\ log x} = \int\ \frac{1}{log x}\ \frac{1}{x}\ dx\]

Put      u  =     log x

\[\frac{du}{dx}=\frac{1}{x}\]
\[du = \frac{1}{x}\ dx\]
\[\int\ \frac{dx}{x\ log x}\ dx= \int\ \frac{du}{u}\]

=  log u + c

=  log ( log x ) + c

\[\int\ \frac{dx}{x\ log x}\ dx= log ( log x) + c\]
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Example:

\[Evaluate\ :\ \int\ \frac{2x + 9}{ x^2 + 9x + 30}\ dx\]

Soln:

\[put\ u\ =\ x^2 + 9x + 30\]
\[\frac{du}{dx}= \ 2x + 9\]
\[du = (2x + 1) dx\]
\[\int\ \frac{2x+ 9}{x^2 + 9x + 30}\ dx= \int\ \frac{du}{u}\]
\[=log\ u + c\]
\[=log( x^2 + 9x + 30) + c\]

Example:

\[Evaluate\ :\ \int\ \frac{sec^2(log x)}{ x}\ dx\]

Soln:

   Put    u  = log x

\[\frac{du}{dx} = \frac{1}{x}\]
\[du = \frac{1}{x}\ dx\]
\[\int\ \frac{sec^2(log x)}{ x}\ dx = \int sec^2\ du\]

=  tan u  + c

=   tan ( log x ) + c

Example:

\[Evaluate\ :\ \int\ sin^3x\ cos x dx\]

Soln:

   Put    u  =  sin x

\[\frac{du}{dx} = cos x\]

du  =  cos x  dx

\[\int\ sin^3x\ cos x dx = \int u^3\ du\]
\[=\frac{u^4}{4} + c\]
\[=\frac{sin^4 x}{4} + c\]

Example:

\[Evaluate\ :\ \int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx\]

Soln:

\[put\ u\ =\ ax^2 + bx + c\]
\[\frac{du}{dx}= \ 2ax + b\]

du  = (2ax + b) dx

\[\int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx= \int\ \frac{du}{\sqrt{u}}\]
\[=\frac{u^{\frac{-1}{2} + 1}}{\frac{-1}{2} + 1} + c\]
\[=\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c\]
\[=2u^{\frac{1}{2}} + c\]
\[=2(ax^2+bx+c)^{\frac{1}{2}}\]