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Monday, 4 August 2014

CHAPTER 14- Miscellaneous Substitutions

In this section, we intend to discuss more advanced techniques of integration and specific substitutions that we’ve not covered till now. 
In the following discussion, the symbol L(x)  represents a linear expression in x (of the form ax + b) while Q(x) represents a quadratic expression (of the form a{x^2} + bx + c). {P_n}(x) would represent a polynomial of degrees n greater than two. M(x) represents a general polynomial. R(a,b,c \ldots) would represent a rational function of the variables a,b,c \ldots.
Recall that we have already developed the requisite techniques to evaluate these types of integrals:
(a) \dfrac{{{M_1}\left( x \right)}}{{{M_2}\left( x \right)}}: ({M_2}\left( x \right) is factorisable into linear or quadratic functions)
If deg ({M_1}(x)) < \deg ({M_2}(x)), we expand this expression using partial dfractions. If ({M_1}(x)) \ge \deg ({M_2}(x)), we first divide {M_1}(x) by {M_2}(x) to obtain the quotient and the remainder and then apply expansion by partial fractions.
(b) \dfrac{1}{{Q\left( x \right)}}:
Depending on the coefficients in Q(x), this integral is of the standard form (17)or (21)
(c) \dfrac{{L\left( x \right)}}{{Q\left( x \right)}}:
Find constants \alpha  and \beta  such that L\left( x \right) = \alpha Q'\left( x \right) + \beta \,\,\,\,Q(x) is not factorisable
(d) \dfrac{1}{{\sqrt {Q\left( x \right)} }}:
Depending on the coefficients in Q(x), this integral is of the standard form (15)(22) or (23)
(e) \dfrac{{L\left( x \right)}}{{\sqrt {Q\left( x \right)} }}:
Find constants \alpha \, and \beta  such that L\left( x \right) = \alpha \,Q'\left( x \right) + \beta
(f) \sqrt {Q\left( x \right)} :
Depending on the coefficients in Q(x), this integral is of the standard form (26)(27) or (28)
(g) L\left( x \right)\sqrt {Q\left( x \right)}:
Find constants \alpha \, and \beta  such that L\left( x \right) = \alpha Q'\left( x \right) + \beta
Let us now consider more forms of this sort. You will observe that the basic unifying theme to solve any integral is the same: we must somehow try to reduce the integral given to us to one of the standard simpler forms.
Consider an expression of the form \dfrac{{{Q_1}\left( x \right)}}{{\sqrt {{Q_2}\left( x \right)} }}. To integrate this, we find constants \alpha ,\beta \, and \gamma  such that
{Q_1}\left( x \right) = \alpha {Q_2}\left( x \right) + \beta {Q_2}'\left( x \right) + \gamma
Thus, this integral becomes
\begin{array}{l}  I = \alpha \int {\dfrac{{{Q_2}\left( x \right)}}{{\sqrt {{Q_2}\left( x \right)} }}} \,\,dx\, + \beta \int {\dfrac{{{Q_2}'\left( x \right)}}{{\sqrt {{Q_2}\left( x \right)} }}} \,dx + \,\gamma \int {\dfrac{1}{{\sqrt {{Q_2}\left( x \right)} }}} \,dx\\  \,\\  \,\,\,\, = \alpha {I_1} \,\,\,\,\,\,\,\, + \,\beta {I_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\, +   \gamma {I_3}  \,\,\,\,\,  \end{array}
{I_1} is actually \int {\sqrt {{Q_2}\left( x \right)} } \,dx which, depending on what the coefficients of{Q_2}(x) are, is one of the standard forms (26)(27) or (28) {I_2} can be evaluated using the substitution {Q_2}(x) = t{I_3} is again one of the standard forms (15)(22) or (23), depending on the coefficients of {Q_2}(x).
Using the same approach, we can evaluate integrals of the form \int {{Q_1}\left( x \right)} \sqrt {{Q_2}\left( x \right)} \,\,dx. If we again express {Q_1}(x) in terms of {Q_2}(x) as described above, this integral becomes
\begin{array}{l}  I = \alpha \int {{Q_2}\left( x \right)\sqrt {{Q_2}\left( x \right)} } \,dx + \beta \int {{Q_2}'\left( x \right)\sqrt {{Q_2}\left( x \right)} } \,dx + \gamma \int {\sqrt {{Q_2}\left( x \right)} } \,\,dx\\  \\  \,\,\,\, = \,\alpha {I_1}  \,\,\,\, + \,\beta \,{I_2}\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\gamma {I_3}  \end{array}
How to evaluate the integrals {I_2} and {I_3} should be obvious. How to evaluate {I_1}is discussed in the following example.
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