ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 14- Miscellaneous Substitutions

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.

Monday, 4 August 2014

CHAPTER 14- Miscellaneous Substitutions

No comments:

https://www.youtube.com/TarunGehlot