Examples On Integration By Substitution Set-11

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Let us now consider rational trignometric functions i.e., functions that involve rational expressions of trignometric terms:

Example – 29

Evaluate the following integrals:

(a) \(\begin{align}\int {\frac{1}{{1 + \sin x}}} \,\,dx\end{align}\)

(b) \(\begin{align}\int {\frac{1}{{1 + \sin x + \cos x}}} \,\,dx\end{align}\)

Solution: In both these examples, the substitutions that we have seen uptill now will not work. (You are urged to try it for yourself).

For such functions, we use a new type of substitution. We use the following half - angle formulae:

\[\begin{align}\sin x = \frac{{2\tan x/2}}{{1 + {{\tan }^2}x/2}}\\ \cos x = \frac{{1 - {{\tan }^2}x/2}}{{1 + {{\tan }^2}x/2}}\end{align}\]

Making these substitutions converts the given integrals into a form where the substitution t = \(\tan \frac{x}{2}\) is possible. Lets apply it on the current example.

(a)                                                                                                \(\begin{align} I = \int {\frac{1}{{1 + \sin x}}} \,\,dx\end{align}\)

\[\begin{align}& = \int {\frac{1}{{1 + \frac{{2\tan x/2}}{{1 + {{\tan }^2}x/2}}}}} \,\,dx\\ &= \int {\frac{{1 + {{\tan }^2}x/2}}{{1 + 2\tan x/2 + {{\tan }^2}x/2}}} \,\,dx\\& = \int {\frac{{{{\sec }^2}x/2}}{{{{\left( {1 + \tan x/2} \right)}^2}}}} \,\,dx\end{align}\]

We now substitute \(\begin{align}\tan \frac{x}{2} = t.\end{align}\) Thus, \(\begin{align}\frac{1}{2}{\sec ^2}\frac{x}{2}dx = dt\end{align}\)

\[\begin{align} &\Rightarrow\quad I = 2\int {\frac{{dt}}{{{{\left( {1 + t} \right)}^2}}}} \\ &\quad\qquad= \frac{{ - 2}}{{1 + t}} + C\\&\quad\qquad = \frac{{ - 2}}{{1 + \tan x/2}} + C\end{align}\]

(b)                                                                                            \(\begin{align} I = \int {\frac{1}{{1 + \sin x + \cos x}}} \,\,dx\end{align}\)

\[\begin{align}& = \int {\frac{1}{{1 + \frac{{2\tan x/2}}{{1 + {{\tan }^2}x/2}} + \frac{{1 - {{\tan }^2}x/2}}{{1 + {{\tan }^2}x/2}}}}} \,\,dx\\ &= \int {\frac{{{{\sec }^2}x/2}}{{2 + 2\tan x/2}}} \,\,dx\end{align}\]

Substitute  \(\begin{align}\tan \frac{x}{2} = t.\end{align}\)Thus,\(\begin{align} {\sec ^2}\frac{x}{2}dx = 2dt\end{align}\)

\[\begin{align}& \Rightarrow\quad I = \int {\frac{{dt}}{{1 + t}}}\\&\qquad\quad = \ln \left| {1 + t} \right| + C\\ &\qquad\quad= \ln \left| {1 + \tan \frac{x}{2}} \right| + C\end{align}\]

These two examples should make it clear that a general integral of the form \(\begin{align} I = \int {\frac{1}{{a\sin x + b\cos x + c}}} \,\,dx\end{align}\) can always be integrated using the mentioned substitutions. Let us analyse the general case itself.

\[\begin{align}& I = \int {\frac{1}{{a\sin x + b\cos x + c}}} \,\,dx...{\rm{ }}\left( 1 \right)\\ & \;\;= \int {\frac{1}{{\frac{{2a\tan x/2}}{{1 + {{\tan }^2}x/2}} + \frac{{b - b{{\tan }^2}x/2}}{{1 + {{\tan }^2}x/2}} + c}}} \,\,dx\\ &\;\;= \int {\frac{{{{\sec }^2}x/2}}{{\left( {c - b} \right){{\tan }^2}x/2 + 2a\tan x/2 + \left( {c + b} \right)}}} \,\,dx\end{align}\]

The substitution \(\begin{align}\tan \frac{x}{2} = t\end{align}\) reduces this integral to

\[\begin{align}& I = 2\int {\frac{{dt}}{{\left( {c - b} \right){t^2} + 2at + \left( {c + b} \right)}}}\\ &\;\;= 2\int {\frac{{dt}}{{A{t^2} + Bt + C}}} \end{align}\]

We have already evaluated integrals of this form. They correspond to either (17) or (21).

\[\left\{ \begin{align}&{\text{In}}\left( 1 \right){\text{above, if }}c = 0,\,\,I\,{\text{can also be solved by writing}}\left( {a\sin x + b\cos x} \right) = \sqrt {{a^2} + {b^2}} \,\sin \left( {x + \phi } \right)\\&\left\{ {\phi = {{\tan }^{ - 1}}\frac{b}{a}} \right\}\,\,{\text{so that }}I{\text{ becomes}}\,\,\frac{1}{{\sqrt {{a^2} + {b^2}} }}\int {{\text{cosec}}\left( {x + \phi } \right)dx} \end{align} \right\}\]

Example – 30

Evaluate the following integrals:

 (a) \(\begin{align}\int {\frac{1}{{1 + {{\sin }^2}x + 2{{\cos }^2}x}}dx}\end{align} \)

(b) \(\begin{align}\int {\frac{1}{{{{\left( {a\sin x + b\cos x} \right)}^2}}}dx} \end{align}\)

Solution: You are first urged to try out the same substitution that we used in the last example, and you’ll see how lengthy the integrals become.

We can solve these integrals more easily by simply dividing the numerator and denominator by \(co{s^2}x\), which converts the given expression into a modified form containing tan x and sec x terms:

(a)                                                                              \(\begin{align} I = \int {\frac{1}{{1 + {{\sin }^2}x + 2{{\cos }^2}x}}dx}\end{align} \)

\[\begin{align}& = \int {\frac{{{{\sec }^2}x}}{{{{\sec }^2}x + {{\tan }^2}x + 2}}dx} & \left\{ \begin{array}{l}{\rm{Now\; write \;the\; se}}{{\rm{c}}^{\rm{2}}}x\\{\rm{in \;the\; denominator}}\\{\rm{as}}\,1 + {\tan ^2}x\end{array} \right\}\\& = \int {\frac{{{{\sec }^2}x}}{{3 + 2{{\tan }^2}x}}dx} \end{align}\]

Substituting tan x = t reduces this integral to

\[ I = \int {\frac{{dt}}{{3 + 2{t^2}}}}\\ \qquad\qquad\; = \frac{1}{2}\int {\frac{{dt}}{{{{\left( {\sqrt {\frac{3}{2}} } \right)}^2} + {t^2}}}} \]

which is of the form (17)

(b)                                                                      \(\begin{align} I = \int {\frac{1}{{{{\left( {a\sin x + b\cos x} \right)}^2}}}} \,\,dx\end{align}\)

\[\begin{align}& = \int {\frac{{{{\sec }^2}x}}{{{{\left( {a\tan x + b} \right)}^2}}}} \,\,dx\\& = \int {\frac{{dt}}{{{{\left( {at + b} \right)}^2}}}} \,\, & \,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{Substituting}}\tan x = t} \right\}\\& = \frac{{ - 1}}{{a\left( {at + b} \right)}} + C\\ &= \frac{{ - 1}}{{a\left( {a\tan x + b} \right)}} + C\end{align}\]

All integrals of the general form \(\begin{align} I = \int {\frac{1}{{a + b{{\sin }^2}x + c{{\cos }^2}x}}} \,\,dx\;or\;I = \int {\frac{1}{{{{\left( {a\sin x + b\cos x} \right)}^2}}}} \,\,dx\end{align}\) can therefore be solved using the approach described above.

We now finally consider another general form of rational trigonometric functions.

Example – 31

\(\begin{align} {\rm{Evaluate}}\,\,\int {\frac{{a\sin x + b\cos x + c}}{{d\sin x + e\cos x + f}}} \,\,dx\end{align}\)

Solution: First suppose that \(c = f = 0.\) In that case, we can find constants \(\lambda \,{\rm{and}}\,\,\mu \) such that the numerator \(\left( {a\sin x + b\cos x} \right)\) can be expressed as a linear combination of the denominator \(\left( {d\sin x + e\cos x} \right)\) and its derivative, using the constants \(\lambda \,{\rm{and}}\,\,\mu \) , i.e:

\[\begin{align}& a\sin x + b\cos x = \lambda \left( {d\sin x + e\cos x} \right) + \mu \left( {d\sin x + e\cos x} \right)'\\&\qquad\qquad\qquad\;\; = \left( {\lambda d - \mu e} \right)\sin x + \left( {\lambda e + \mu d} \right)\cos x\end{align}\]

(convince yourself that we can always find such constants \(\lambda \,\,{\rm{and}}\,\,\mu \) )

Comparing the coefficients of sin x and cos x on both sides will uniquely give us the constants \(\lambda \,\,{\rm{and}}\,\,\mu \)

\[\begin{array}{l}\lambda d - \,\mu e = a\\ \lambda e + \,\mu d = b\end{array}\]

Having found these constants, our integral reduces to

\[\begin{align}& I = \lambda \int {dx + \mu \int {\frac{{\left( {d\sin x + e\cos x} \right)'}}{{d\sin x + e\cos x}}} \,\,dx} \\&\;\; = \lambda x + \mu {I_1}\end{align}\]

\({I_1}\) can simply be solved by the substitution \(\left( {d\sin x + e\cos x} \right) = t\) which reduces \(\begin{align} {I_1}\,\,{\rm{to}}\,\,\int {\frac{{dt}}{t} = \ln \left| t \right|} = \ln \left| {d\sin x + e\cos x} \right|\end{align}\)

Now suppose that at least one of c and f is non-zero. In this case, we can (always) find three constants \(\lambda ,\mu \,\,{\rm{and}}\,\,\theta \) such that

\[\left( {{\rm{Numerator}}} \right) = \lambda \left( {{\rm{Denominator}}} \right) + \mu \left( {{\rm{Denominator}}} \right)' + \theta \]


\[\left( {a\sin x + b\cos x + c} \right) = \lambda \left( {d\sin x + e\cos x + f} \right) + \mu \left( {d\cos x - e\sin x} \right) + \theta \\ \qquad\qquad\qquad\; = \left( {\lambda d - \mu e} \right)\sin x + \left( {\lambda e + \mu d} \right)\cos x + \lambda f + \theta \]

Thus, \(\lambda ,\mu \,\,{\rm{and}}\,\,\theta \) can be determined using

\[\begin{array}{l}\lambda d - \mu e = a\\ \lambda e + \mu d = b\\ \lambda f + \theta = c\end{array}\]

Our integral now reduces to

\[\begin{align} & I = \lambda \int {dx + \mu \int {\frac{{d\cos x - e\sin x}}{{d\sin x + e\cos x + f}}dx + \theta \int {\frac{1}{{d\sin x + e\cos x + f}}} \,dx} } \\&\;\;= \lambda x + \mu {I_2} + \theta {I_3}\end{align} \]

\({I_2}\) can be evaluated by the simple substitution \(\left( {d\sin x + e\cos x + f} \right) = t.\) To evaluate \({I_3}\) , we follow the approach already described in example - 29

You should not feel confused by the wide variety of integrals that we’ve encountered upto this point. We would strictly advise you against memorizing the solutions to all the different types of integrals. What you need to do is remember the unifying technique of substitution and how it is applied to different cases. It is possible that in the exam, you might encounter an integral of a form you’ve never seen before. Thus, it would be of no use memorizing all the results (if at all, you are able to do so). You should instead, focus on learning the technique and not the particular results.

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Indefinite Integration
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