Advanced Examples On Integration Set-7

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We now need to deal with one final form of integrals. Integrals that can be characterised by an integer n can be evaluated by reduction formulae (you can also call them recursive relations).

Suppose we have an integral \({I_n}\) that we wish to evaluate. As an example, think about \({I_n} = {\sin ^n}x\,\,dx\) . The order of the integral \({I_n}\) is n. What we want to do is write a reduction formula (a recursive relation) which relates \({I_n}\) to a lower order integral, say for example \({I_{n - 1}}\) . If we have in \({I_n}\) terms of \({I_{n - 1}}\) in this relation, we can use the same relation to evaluate \({I_{n - 1}}\) in terms of \({I_{n - 2}}\) , which could be expressed in terms of \({I_{n - 3}}\) , and so on, all the way upto, say for example, \({I_1}{\rm{ or }}{I_0}\) which could easily be evaluated. This is the whole point of writing a reduction formula. Using such a formula, we try to relate a higher order and a lower order integral.

The technique of writing such relations is elaborated in the following examples.

Example – 59

Write reduction formulae for the following integrals

(a) \(\int {{{\sin }^n}x\,\,dx} \)

(b) \(\int {{{\cos }^n}x\,\,dx} \)

(c) \(\int {{{\tan }^n}x\,\,dx} \)

(d) \(\int {{{\cot }^n}x\,\,dx} \)

(e) \(\int {{{\sec }^n}x\,\,dx} \)

(f) \(\int {{\rm{cose}}{{\rm{c}}^n}x\,\,dx} \)

Solution: (a)                                                           \({I_n} = \int {{{\sin }^n}x\,\,dx} \)

\[\begin{align}&\,\,\,\,\,\, \qquad= \int {\mathop {{{\sin }^{n - 1}}x}\limits_{{\rm{Ist}}} \cdot \mathop {\sin x}\limits_{{\rm{IInd}}} \,\,dx} \\\\&\,\,\,\,\,\, \qquad= - \cos x \cdot {\sin ^{n - 1}}x - (n - 1)\int {{{\sin }^{n - 2}}x \cdot \cos x \cdot ( - \cos x)\,\,dx} \\\\&\,\,\,\,\,\,\qquad = - \cos x{\sin ^{n - 1}}x + (n - 1)\int {{{\sin }^{n - 2}}x(1 - {{\sin }^2}x)\,\,dx} \\\\&\,\,\,\,\,\,\qquad = - \cos x{\sin ^{n - 1}}x + (n - 1)\int {{{\sin }^{n - 2}}x\,dx - \left( {n - 1} \right)\int {{{\sin }^n}x\,dx} } \\\\&\,\,\,\,\,\,\qquad = - \cos x{\sin ^{n - 1}}x + (n - 1){I_{n - 2}} - \left( {n - 1} \right){I_n}\\\\ &\Rightarrow \quad {I_n} = - \frac{{\cos x{{\sin }^{n - 1}}x}}{n} + \frac{{\left( {n - 1} \right)}}{n}{I_{n - 2}}\end{align}\]

Observe that this reduction formula relates  \({{\text{I}}_{\text{n}}}\text{ to }{{\text{I}}_{\text{n-2}}}\text{.}\) . Using exactly the same approach, we obtain a similar relation for our next part

(b) \(\begin{align}{I_n} = \int {{{\cos }^n}xdx = \frac{{{{\cos }^{n - 1}}x\sin x}}{n}} + \frac{{\left( {n - 1} \right)}}{n}{I_{n - 2}}\end{align}\)

(c)                                                                                                  \({I_n} = \int {{{\tan }^n}xdx} \)

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

Thus,

\(\begin{align}{I_n} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - {I_{n - 2}}\end{align}\)

(d) Using an analogous approach as in the part above, we obtain the required relation for this part

\[{I_n} = \int {{\rm{co}}{{\rm{t}}^n}} xdx = - \frac{{{{\cot }^{n - 1}}x}}{{n - 1}} - {I_{n - 2}}\]

(e)                                                                                 \({I_n} = \int {{{\sec }^n}xdx} \)

\[\begin{align}&\qquad\quad\; =\int{{{\sec }^{n-2}}x\cdot {{\sec }^{2}}xdx}  \\&\qquad\quad\;\qquad\qquad \text{Ist}\quad\text{IInd}  \\
&\qquad\quad\;=\tan x\cdot {{\sec }^{n-2}}x-\left( n-2 \right)\int{{{\sec }^{n-2}}x\tan xdx}  \\
&\qquad\quad\;=\tan x\cdot {{\sec }^{n-2}}x-\left( n-2 \right)\int{{{\sec }^{n-2}}x\left( {{\sec }^{2}}x-1 \right)dx}  \\
 &\qquad\quad\;  =\tan x\cdot {{\sec }^{n-2}}x-\left( n-2 \right)\int{{{\sec }^{n}}xdx+\left( n-2 \right)\int{{{\sec }^{n-2}}xdx}}  \\
 \\  & \Rightarrow \quad {{I}_{n}}=\frac{\tan x\cdot {{\sec }^{n-2}}x}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}} \\ 
\end{align}\]

Using exactly the same approach, we obtain a similar relation for the next part.

(f) \(\begin{align}{I_n} = \int {{\rm{cose}}{{\rm{c}}^n}xdx} = - \frac{{\cot x{\rm{cose}}{{\rm{c}}^{n - 2}}x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}{I_{n - 2}}\end{align}\)

Example – 60

Find a reduction formula for the integral \({I_{m,n}} = \int {{e^{mx}}{x^n}dx} \)

Solution: \[\begin{align}{I_{m,n}} = \int {{x^n} \cdot {e^{mx}}dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Ist}}\,\,\,\;{\rm{IInd}}\,\end{align}\]

\[\begin{align}& = \frac{{{x^n} \cdot {e^{mx}}}}{m} - \frac{n}{m}\int {{x^{n - 1}}{e^{mx}}dx} \\\\& = \frac{{{x^n}{e^{mx}}}}{m} - \frac{n}{m}{I_{m,\,n - 1}}\end{align}\]

This relation relates \({I_{m,\,n}}\,\,{\rm{to}}\,\,{I_{m,\,\,n - 1}}\) , which is lower in index (considering n). Thus, this is a valid recursive relation

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Indefinite Integration
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