Examples On Integration By Substitution Set-7

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Example - 22

Evaluate the following integrals:

(a) \(\int {{{\sin }^3}x{{\cos }^4}xdx} \) (b) \(\int {{{\sin }^2}x{{\cos }^5}x\,\,dx} \)
(c) \(\int {{{\sin }^3}x} {\cos ^5}x\,dx\) (d) \(\int {{{\cos }^5}xdx} \)
(e) \(\begin{align}\int {\frac{1}{{\sqrt[4]{{{{\sin }^3}x{{\cos }^5}x}}}}dx} \end{align}\) (f) \(\int {{{\sin }^4}x{{\cos }^2}x\,dx} \)
(g) \(\int {{{\cos }^8}x\,dx} \)

(h) \(\int {{{\sec }^4}x\,\,{\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx} \)

 

Solution: (a) \(I = \int {{{\sin }^3}x{{\cos }^4}x\,dx} \)

Notice that the power of \(\sin x\) is odd while that of \(\cos x\) is even. What substitution can we possibly use? If we use \(\sin x = t,\) we will get \(\cos x\,dx = dt.\) Thus, we’ll be left with \(\int {{t^3}{{\cos }^3}x\,dt,} \) which contains \({\cos ^3}x;\) this will be tedious to express in terms of \(t(\sin x).\) We therefore use the substitution \(\cos x = t;\) you will soon see that this is an appropriate substitution.

\[\begin{align}&\qquad\;\;\cos x = t \\ &\Rightarrow\quad \sin x\,dx = - dt\\ &\Rightarrow\quad I = \int {{{\sin }^3}x{{\cos }^4}x\,dx} \\ &\qquad\quad\;= - \int {{{\sin }^2}x \cdot {t^4}dt} \\ &\qquad\quad\;= - \int {(1 - {t^2}){t^4}dt} \\ &\qquad\quad\; = \int {({t^6}} - {t^4})dt\\ &\qquad\quad\; = \frac{{{t^7}}}{7} - \frac{{{t^5}}}{5} + C\\ &\qquad\quad\; = \frac{{{{\cos }^7}x}}{7} - \frac{{{{\cos }^5}x}}{5} + C\end{align}\]

(b) \(I = \int {{{\sin }^2}x{{\cos }^5}x\,dx} \)

If you compare this integral with the one in part-(a), you’ll see that the roles have been reversed. \(\sin x\) has an even power while the power of \(\cos x\) is odd. We therefore use the substitution \(\sin x = t\)

\[\begin{align}&\Rightarrow \quad  \cos x\,dx = dt\\& \Rightarrow\quad I = \int {{t^2}{{\cos }^4}x\,dt}\\&\quad\qquad= \int {{t^2}{{(1 - {t^2})}^2}dt} \\ & \quad\qquad= \int {{t^2}({t^4} - 2{t^2} + 1)dt} \\&\quad\qquad = \int {({t^6} - 2{t^4} + {t^2})dt} \\ &\quad\qquad = \frac{{{t^7}}}{7} - \frac{{2{t^5}}}{5} + \frac{{{t^3}}}{3} + C\\ &\quad\qquad = \frac{{{{\sin }^7}x}}{7} - \frac{{2{{\sin }^5}x}}{5} + \frac{{{{\sin }^3}x}}{3} + C\end{align}\]

(c) \(I = \int {{{\sin }^3}x{{\cos }^5}x\,dx} \)

Here, the powers of \(\sin x\;and\;\cos x\) are both odd. This suggests that we can substitute either \(\sin x\;or\;\cos x\) as a new variable. We use \(\sin x = t.\)

\[\begin{align}&\Rightarrow \quad\cos x\,dx = dt\\&\Rightarrow \quad I= \int {{t^3}{{\cos }^4}x\,dt}\\ &\quad\qquad= \int {{t^3}{{(1 - {t^2})}^2}dt} \\ & \quad\qquad= \int {{t^3}({t^4} - 2{t^2} + 1)dt} \\& \quad\qquad= \int {({t^7} - 2{t^5} + {t^3})dt} \\&\quad\qquad = \frac{{{t^8}}}{8} - \frac{{{t^6}}}{3} + \frac{{{t^4}}}{4} + C\\&\quad\qquad= \frac{{{{\sin }^8}x}}{8} - \frac{{{{\sin }^6}x}}{3} + \frac{{{{\sin }^4}x}}{4} + C\end{align}\]

(d) \(I = \int {{{\cos }^5}x\,dx} \)

The power of \(\cos x\) is odd. As in part-(b), we use the substitution \(\sin x = t\) \[\begin{align} &\Rightarrow \quad\cos x\,dx = dt\\ &\Rightarrow \quad I = \int {{{\cos }^4}x\,dt} \\  &\quad\qquad= \int {{{(1 - {t^2})}^2}dt} \\& \quad\qquad= \int {({t^4} - 2{t^2} + 1)dt} \\& \quad\qquad= \frac{{{t^5}}}{5} - \frac{{2{t^3}}}{3} + t + C\\ &\quad\qquad= \frac{{{{\sin }^5}x}}{5} - \frac{{2{{\sin }^3}x}}{3} + \sin x + C\end{align}\]

(e)\(\begin{align}I = \int {\frac{1}{{\sqrt[4]{{{{\sin }^3}x{{\cos }^5}x}}}}dx} \end{align}\)

Observe that neither the substitution on \(\sin x = tor\cos x = t\) would help. Some rearrangement needs to be done before we can substitute for something

\[\begin{align}& I = \int {\frac{1}{{\sqrt[4]{{{{\sin }^3}x{{\cos }^5}x}}}}dx}\\ &\;\;= \int {\frac{1}{{\sqrt[4]{{{{\cos }^8}x \cdot \frac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}}}dx} & \left\{ \begin{gathered}{\text{How did we hit upon this rearrangement? }}\\{\text{Answer:We are trying to rearrange the }}\\{\text{expression so that it contains}}\tan x{\text{and}}\\\sec  x{\text{ terms}}\end{gathered} \right\}\\ &\;\;= \int {\frac{1}{{{{\cos }^2}x\sqrt[4]{{{{\tan }^3}x}}}}} dx\\ &\;\;= \int {\frac{{{{\sec }^2}x}}{{\sqrt[4]{{{{\tan }^3}x}}}}dx} \end{align}\]

Now we use the substitution \(\tan x = t\)

\[\begin{array}{l} \Rightarrow\quad {\sec ^2}x\,dx = dt\\ \Rightarrow \quad I = \int {\frac{{dt}}{{{t^{3{\rm{/}}4}}}}} \end{array}\]

\[\begin{array}{l}\qquad\qquad\qquad = 4{t^{1{\rm{/}}4}} + C\\ \qquad\qquad\qquad= 4{(\tan x)^{1{\rm{/}}4}} + C\end{array}\]

This example is one where the powers of \(\sin x\) and \(\cos x\) add up to a negative integer

\(\left\{ {{\rm{in \;this\; question}},\left( { - \frac{3}{4}} \right) + \left( { - \frac{5}{4}} \right) = - 2} \right\}\)

(f) \(I = \int {{{\sin }^4}x{{\cos }^2}x\,dx} \)

Both the powers are even. Observe that neither of the substitutions \(\sin x = tor\cos x = t\) or would help. Therefore, we again try to convert this expression into another form which contains \(\tan xand\sec x\) and terms, as in the previous examples.

\[\begin{align}& I = \int {{{\sin }^4}x{{\cos }^2}x\,dx} \\&\;\; = \int {{{\cos }^6}x \cdot \frac{{{{\sin }^4}x}}{{{{\cos }^4}x}}dx} \\ &\;\;= \int {\frac{{{{\tan }^4}x}}{{{{\sec }^6}x}}dx} \end{align}\]

Not much headway! What do we do now?

Neither the substitution \(\tan x = tor\sec x = t\) would help much in the final integral above.

We now follow an alternative approach. We use multiple angle formulae to reduce the non-linear trignometric terms in the integral expression to linear trignometric terms.

\[\begin{align}&\quad{\sin ^4}x = {({\sin ^2}x)^2}\\&\qquad\quad\; = {\left( {\frac{{1 - \cos 2x}}{2}} \right)^2}\\ &\qquad\quad\;= \frac{1}{4}\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)\\ &\qquad\quad\;= \frac{1}{4}\left( {1 - 2\cos 2x + \frac{{1 + \cos 4x}}{2}} \right)\\&\qquad\quad\;= \frac{1}{8}\left( {3 - 4\cos 2x + \cos 4x} \right)\end{align}\]

Similarly, \(\begin{align}{\cos ^2}x = \frac{{1 + \cos 2x}}{2}\end{align}\)

\[\begin{align}&\Rightarrow\quad I = \frac{1}{{16}}\int {(1 + \cos 2x)} (3 - 4\cos 2x + \cos 4x)dx\\& \quad\qquad= \frac{1}{{16}}\int {\left\{ {3 - \cos 2x + \cos 4x - 4{{\cos }^2}2x + \cos 2x\cos 4x} \right\}\,} dx\\ & \quad\qquad= \frac{1}{{16}}\int {\left\{ {3 - \cos 2x + \cos 4x - \frac{{4(1 + \cos 4x)}}{2} + \frac{{\cos 6x + \cos 2x}}{2}} \right\}} \,dx\\ &\quad\qquad= \frac{1}{{16}}\int {\left\{ {1 - \frac{1}{2}\cos 2x - \cos 4x + \frac{1}{2}\cos 6x} \right\}\,} dx\\& \quad\qquad= \frac{1}{{16}}\left( {x - \frac{{\sin 2x}}{4} - \frac{{\sin 4x}}{4} + \frac{{\sin 6x}}{{12}}} \right) + C\end{align}\]

(g) \(I = \int\limits_{}^{} {{{\cos }^8}x\,\,dx} \)

The power of cos x is even . As in the previous example, we use multiple angle formulae to reduce the non-linear trignometric term \(({\cos ^8}x)\) to (a combination of) linear trignometric terms.

 \[\begin{align}I &= \int\limits_{}^{} {{{\cos }^8}x\,\,dx}\\&= \int {{{\left( {\frac{{1 + \cos 2x}}{2}} \right)}^4}\,\,dx} \\ &= \frac{1}{{16}}\int {{{\left( {{{(1 + \cos 2x)}^2}} \right)}^2}\,\,dx}\\ &= \frac{1}{{16}}\int {{{\left( {1 + 2\cos 2x + {{\cos }^2}2x} \right)}^2}\,\,dx}\\&= \frac{1}{{16}}\int {{{\left( {1 + 2\cos 2x + \frac{{1 + \cos 4x}}{2}} \right)}^2}\,\,dx} \\&= \frac{1}{{64}}\int {{{\left( {3 + 4\cos 2x + \cos 4x} \right)}^2}\,\,dx} \\&= \frac{1}{{64}}\int {\left\{ {9 + 16{{\cos }^2}2x + {{\cos }^2}4x + 24\cos 2x + 6\cos 4x + 8\cos 2x\cos 4x} \right\}} \\&= \frac{1}{{64}}\int {\left\{ {9 + 16\left( {\frac{{1 + \cos 4x}}{2}} \right) + \frac{{1 + \cos 8x}}{2} + 24\cos 2x + 6\cos 4x + 4\cos 6x + 4\cos 2x} \right\}\,\,dx} \\&= \frac{1}{{64}}\int {\left\{ {\frac{{35}}{2} + 28\cos 2x + 14\cos 4x + 4\cos 6x + \frac{1}{2}\cos 8x} \right\}\,\,dx} \end{align}\]   

This expression involves purely linear trignometric terms and can easily be integrated. Thus,

\[I = \frac{{35}}{{128}}x + \frac{7}{{32}}\sin 2x + \frac{7}{{128}}\sin 4x + \frac{1}{{96}}\sin 6x + \frac{1}{{1024}}\sin 8x + C\]

You may rest assured that you’ll never encounter such a lengthy question in an actual exam.This  example was just included to illustrate the techniques involved.

(h) \(\begin{align} I = \int {{{\sec }^4}x\,\,{\rm{cose}}{{\rm{c}}^2}x\,\,dx = \int {\frac{1}{{{{\sin }^2}x\,\,{{\cos }^4}x}}\,\,dx} } \end{align}\)

Again, neither the substitution \(\sin x = tor\cos x = t\) or would help. We also cannot use multiple angle formulae as we’ve done in the previous two examples. However, as in part -(e), we can convert this expression into one involving \(\tan x\;and\;\sec x\) terms.

\[\begin{align}&I = \int {\frac{1}{{{{\sin }^2}x{{\cos }^4}x}}\,\,dx} \\&\;\;= \int {\frac{1}{{{{\tan }^2}x\,\,{{\cos }^6}x}}\,\,dx} \\&\;\; = \int {\frac{{{{\sec }^6}x}}{{{{\tan }^2}x\,\,}}\,\,dx} \end{align}\]

Now, we use the substitution \(\tan x = t\) . Thus, \({\sec ^2}x\,\,dx = dt\)

\[\begin{align}  &\Rightarrow\quad I = \int {\frac{{{{\sec }^4}x}}{{{t^2}}}\,\,dt}\\&\quad\qquad = \int {\frac{{{{(1 + {{\tan }^2}x)}^2}}}{{{t^2}}}\,\,dt} \\&\quad\qquad= \int {\frac{{{{(1 + {{\mathop{\rm t}\nolimits} ^2})}^2}}}{{{t^2}}}\,\,dt} \\ &\quad\qquad= \int {\left( {\frac{{{{\mathop{\rm t}\nolimits} ^4} + 2{t^2} + 1}}{{{t^2}}}} \right)\,\,dt} \,\\ &\quad\qquad= \int {({t^2} + 2 + {t^{ - 2}})\,\,dt} \\ &\quad\qquad= \frac{{{t^3}}}{3} + 2t - \frac{1}{t} + C\\ &\quad\qquad= \frac{{{{\tan }^3}x}}{3} + 2\tan x - \frac{1}{{\tan x}} + C\end{align}\]

Note that like part -(e), this example is also one where the powers of sin x and cos x and up to a negative integer.

From these eight examples, we can make certain generalizations as to how to solve integrals of the form \(I = \int {{{\sin }^m}x{{\cos }^n}x\,\,dx} \) , where \(m,\,\,n \in Q\) .

\(\text{Condition} \) \(\text{Substitution /Action} \)
\(\begin{align}
  & \text{(1)}\;\;m\text{ is an odd integer, }n\,\,\text{is an even integer} \\ 
 &  \\ 
 & \text{(2)}\;\;n\text{ is an odd integer, }m\text{ is an even integer} \\ 
 &  \\ 
 & \text{(3)}\;\;\text{Both }m\text{, }n\text{ are odd integers} \\ 
 &  \\ 
 & \text{(4)}\;\;\text{Both }m\text{, }n\text{ are even integers} \\ 
 &  \\ 
 &  \\ 
 & \text{(5)}\;\;m\text{, }n\text{ add up to a negative integer} \\ 
 &  \\ 
\end{align}\)
\( \begin{align}
  & \cos x=t \\ 
 &  \\ 
 & \sin x=t \\ 
 &  \\ 
 & \sin x=t\,\,\,\text{or}\cos x=t \\ 
 &  \\ 
 & \text{Use multiple-angle formulae to convert the given expression} \\ 
 & \text{into one involving only linear trignometric terms}\text{.} \\ 
 &  \\ 
 & \text{Convert the given expression into one involving tan }x\text{ and } \\ 
 & \text{sec }x\text{ terms and then use the substitution tan }x\text{ = }t \\ 
\end{align}\)

You are urged to draw up a similar table for integrals of the form \(I=\int{{{\tan }^{m}}x{{\sec }^{n}}x\,\,dx,}~where\;m,\,\,n\in Q\) .

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