Methods of Integration

If the integrand is not a derivative of a known function, then the corresponding integrals cannot be found directly. In order to find the integral of complex problems, there are three fundamental methods of integration which are used widely:

INTEGRATION BY SUBSTITUTION

INTEGRATION BY PARTS

INTEGRATION BY PARTIAL FRACTIONS

Integration by Substitution

Direct Substitution:

If the integrand consists of a product of two factors one which contains a part of which the second factor is either an exact derivative or can be made so by multiplying by a suitable constant, then that part of the first factor may be substituted by t.

Illustration 7:

Evaluate $\int_{}^{}\ \frac{dx}{2\sqrt{x}(1 + x)}$

Solution:

$\ I = \int_{}^{}\ \frac{dx}{2\sqrt{x}(1 + x)}$ , put $\sqrt{x} = t \Rightarrow \frac{1}{2\sqrt{x}}dx = dt$

$\therefore I = \int_{}^{}\ \frac{dt}{1 + t^{2}} = \tan^{- 1}t + c = \tan^{- 1}\sqrt{x} + c$

Illustration 8:

Evaluate $\int_{}^{}\ \frac{x^{7}dx}{\left( 1 + x^{2} \right)^{5}}$

Solution:

$\begin{matrix} & I = \int_{}^{}\ \frac{x^{7}dx}{\left( 1 + x^{2} \right)^{5}} = \int_{}^{}\ \frac{x^{7}dx}{x^{10}\left( 1 + \frac{1}{x^{2}} \right)^{5}} = \int_{}^{}\ \frac{dx}{x^{3}\left( 1 + \frac{1}{x^{2}} \right)^{5}}\ \text{Let~}t = 1 + \frac{1}{x^{2}} \Rightarrow dt = - \frac{2dx}{x^{3}} \\ & \ \Rightarrow I = - \frac{1}{2}\int_{}^{}\ \frac{dt}{t^{5}} = \frac{1}{8t^{4}} + c = \frac{1}{8}\left( 1 + \frac{1}{x^{2}} \right)^{- 4} + c \end{matrix}$

Illustration 9:

Evaluate $\int_{}^{}\ \frac{(x + 1)}{x\left( 1 + xe^{x} \right)^{2}}dx$ .

Solution:

$\begin{matrix} & \ \int_{}^{}\ \frac{(x + 1)e^{x}}{xe^{x}\left( 1 + xe^{x} \right)^{2}}dt\ \text{put~}1 + xe^{x} = t,\ (x + 1)e^{x}dx = dt \\ & \ = \int_{}^{}\ \frac{dt}{(t - 1)t^{2}}\ \text{put~}t = \frac{1}{u}dt = - \frac{1}{u^{2}}du \\ & \ \int_{}^{}\ \frac{- \frac{1}{u^{2}}du}{\left( \frac{1 - u}{u} \right)\frac{1}{u^{2}}} = \int_{}^{}\ \frac{- u}{(1 - u)}du \Rightarrow u + ln(1 - u) + c \Rightarrow \frac{1}{1 + xe^{x}} + ln\left( \frac{xe^{x}}{1 + xe^{x}} \right) + c \end{matrix}$

Indirect Substitution

S. NO.	FORM OF INTEGRAL	SUBSTITUTION
1.	For terms of the form $x^{2} + a^{2}$ or $\sqrt{x^{2} + a^{2}}$	Put $x = atan\theta$ or $acot\theta$
2.	For terms of the form $x^{2} - a^{2}$ or $\sqrt{x^{2} - a^{2}}$	Put $x = asec\theta$ or $acosec\theta$
3.	For terms of the form $a^{2} - x^{2}$ or $\sqrt{a^{2} - x^{2}}$	Put $x = asin\theta$ or $x = acos\theta$
4.	If both $\sqrt{a + x},\ \sqrt{a - x}$ are present	put $x = acos\theta$
5.	For the type $\sqrt{(x - a)(b - x)}$ ,	put $x = a\cos^{2}\theta + b\sin^{2}\theta$
6.	For the type $\left( \sqrt{x^{2} + a^{2}} \pm x \right)^{n}$ or $\left( x \pm \sqrt{x^{2} - a^{2}} \right)^{n}$	put the expression within the bracket $= t$ .
7.	For the type $(x + a)^{- 1 - \frac{1}{n}}(x + b)^{- 1 + \frac{1}{n}}$ or $\left( \frac{x + b}{x + a} \right)^{\frac{1}{n} - 1}\frac{1}{(x + a)^{2}},\ (n \in N,\ n > 1)$	put $\frac{x + b}{x + a} = t$ .
8.	For $\frac{1}{(x + a)^{n_{1}}(x + b)^{n_{2}}};n_{1}n_{2} \in \text{ }N(and > 1)$	put $(x + a) = t(x + b)$

Illustration 10:

Evaluate $\int_{}^{}\ \sin^{- 1}\sqrt{\frac{x}{a + x}}dx$

Solution:

Let $x = a\tan^{2}\theta,\ dx = 2atan\theta\sec^{2}\theta\text{ }d\theta$

$I = \int\sin^{- 1}\left( \sqrt{\frac{{atan}^{2}\theta}{{asec}^{2}\theta}} \right)2atan\theta\sec^{2}\theta d\theta$

$I = 2a\int\theta tan\theta\sec^{2}\theta\text{ }d\theta = 2a\left\lbrack \frac{\theta\tan^{2}\theta}{2} - \int\frac{\tan^{2}\theta}{2}\text{ }d\theta \right\rbrack = a\theta\tan^{2}\theta - tan\theta + a\theta$

Writing $\theta$ in terms of x, we get

$I = x\tan^{- 1}\sqrt{\frac{x}{a}} - \sqrt{ax} + x\tan^{- 1}\sqrt{\frac{x}{a}} + c$

Illustration 11:

Evaluate $\int\frac{dx}{\left( a^{2} + x^{2} \right)^{3/2}}$ .

Solution:

$I = \int\frac{dx}{\left( a^{2} + x^{2} \right)^{3/2}}$

Put $x = atan\theta \Rightarrow dx = a\sec^{2}\theta d\theta$

$\begin{matrix} & \ \therefore I = \int_{}^{}\ \frac{{asec}^{2}\theta d\theta}{\left( a^{2} + a^{2}\tan^{2}\theta \right)^{3/2}} = \int_{}^{}\ \frac{{asec}^{2}\theta d\theta}{a^{3}\left( \sec^{2}\theta \right)^{3/2}} = \frac{1}{a^{2}}\int_{}^{}\ \frac{\sec^{2}\theta d\theta}{\sec^{3}\theta} \\ & \ \Rightarrow I = \frac{1}{a^{2}}\int_{}^{}\ \frac{d\theta}{sec\theta} = \frac{1}{a^{2}}\int_{}^{}\ cos\theta d\theta \\ & \ \Rightarrow I = \frac{1}{a^{2}}sin\theta + C \Rightarrow \frac{x}{a^{2}\left( x^{2} + a^{2} \right)^{1/2}} + C \end{matrix}$

Illustration 12:

Evaluate $\int_{}^{}\ \sqrt{\frac{1 - x}{1 + x}}dx$

Solution:

Put $x = cos2\theta$ , then $\theta = \frac{1}{2}\cos^{- 1}x \Rightarrow dx = - 2sin2\theta d\theta$

$\begin{matrix} & \ \therefore I = - 2\int_{}^{}\ \sqrt{\frac{1 - cos2\theta}{1 + cos2\theta}} \cdot sin2\theta d\theta \\ & \ \Rightarrow I = - 2\int_{}^{}\ \sqrt{2\sin^{2}\theta/2\cos^{2}\theta} \cdot sin2\theta \cdot d\theta \\ & \ \Rightarrow I = - 2\int_{}^{}\ tan\theta \cdot 2sin\theta cos\theta d\theta \Rightarrow I = - 2 \cdot 2\int_{}^{}\ \sin^{2}\theta d\theta \Rightarrow I = - 2\int_{}^{}\ (1 - cos2\theta)d\theta \\ & \ \Rightarrow I = - 2\left\lbrack \theta - \frac{sin2\theta}{2} \right\rbrack + C_{1} \Rightarrow I = - 2\theta + sin2\theta + C_{1} \\ & \ \Rightarrow I = - \cos^{- 1}x + \sqrt{1 - x^{2}} + C_{1} \\ & \ \Rightarrow I = - \frac{\pi}{2} + \sin^{- 1}x + \sqrt{1 - x^{2}} + C_{1} \\ & \ \Rightarrow I = \sin^{- 1}x + \sqrt{1 - x^{2}} + C\ \left\lbrack \text{~where~}C = C_{1} - \frac{\pi}{2} \right\rbrack \end{matrix}$

Integration by Parts

If $f(x)$ and $g(x)$ be two functions of $x$ , then integral of product of these two functions is given by: $\int_{}^{}\ \underset{I}{f(x)}\underset{II}{g}(x)dx = f(x)\int_{}^{}\ g(x)dx - \int_{}^{}\ \left\lbrack \frac{df(x)}{dx}\int_{}^{}\ g(x)dx \right\rbrack dx$

Fundamental logics for the selection of first and the second function:

If in the product of the two functions, one of the functions is not directly integrable (e.g. $lnx,\ \sin^{- 1}x,\ \cos^{- 1}x,\ \tan^{- 1}x$ etc.) then we take it as the first function and the remaining function is taken as the second function. e.g. In the integration of $\int_{}^{}\ x\tan^{- 1}xdx,\ \tan^{- 1}x$ is taken as the first function and x as the second function.

If there is no other function, then unity is taken as the second function e.g. In the integration of $\int_{}^{}\ \tan^{- 1}xdx$ , $\tan^{- 1}x$ is taken as the first function and 1 as the second function.

If both of the function are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. For this we follow the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent) In the above stated order, the function on

the left is always chosen as the first function. e.g. In the integration of $\int_{}^{}\ xsinxdx,\ x$ is taken as the first function and $sinx$ is taken as the second function.

In the integral $\int_{}^{}\ g(x)e^{x}dx$ , if $g(x)$ can be expressed as

$g(x) = f(x) + f^{'}(x)\text{~then~}\int_{}^{}\ e^{x}\left\lbrack f(x) + f^{'}(x) \right\rbrack dx = e^{x}f(x) + c$

IIIustration13:

If $I_{n} = \int_{}^{}\ (logx)^{n}dx$ , then find $I_{n} + nI_{n - 1}$ .

Solution:

$I_{n} = \int_{}^{}\ (logx)^{n}dx,\ \therefore I_{n - 1} = \int_{}^{}\ (logx)^{n - 1}dx$

$\begin{matrix} & \text{~Now~}I_{n} = \int_{}^{}\ (logx)^{n} \cdot 1dx \\ & \ = (logx)^{n} \cdot x - n\int_{}^{}\ (logx)^{n - 1} \cdot \frac{1}{x} \cdot xdx \\ & \ = (logx)^{n} \cdot x - n\int_{}^{}\ (logx)^{n - 1}dx \\ & I_{n} = x(logx)^{n} - nI_{n - 1}\therefore I_{n} + nI_{n - 1} = x(logx)^{n} \end{matrix}$

Illustration14:

If $\int_{}^{}\ xsinxdx = - xcosx + A$ , then find $A$ .

Solution:

Since $\int_{}^{}\ xsinxdx = - xcosx + A$

$\Rightarrow - xcosx + sinx + constant = - xcosx + A$

Equating it, we get $A = sinx +$ constant

IIIustration:

Evaluate $\int_{}^{}\ e^{x}\frac{1 + sinx}{1 + cosx}dx$

Solution:

$\begin{matrix} I = & \ \int_{}^{}\ \frac{e^{x}}{1 + cosx}dx + \int_{}^{}\ \frac{e^{x}sinx}{1 + cosx}dx \\ & \ = \int_{}^{}\ \frac{e^{x}}{2\cos^{2}\frac{x}{2}}dx + \int_{}^{}\ \frac{e^{x}2sinx/2 \cdot cosx/2}{2\cos^{2}x/2}dx \\ & \ = \int_{}^{}\ e^{x}\left\lbrack tan\frac{x}{2} + \frac{1}{2}\sec^{2}\frac{x}{2} \right\rbrack dx \\ & \ = e^{x}tan\frac{x}{2} + c = e^{x}\left( f(x) + f^{'}(x) \right) \end{matrix}$

Illustration:

Evaluate $\int_{}^{}\ x\sec^{2}xdx$

Solution:

$\int_{}^{}\ x\sec^{2}xdx = xtanx - \int_{}^{}\ tanxdx = xtanx + log(cosx) + c$

Integration by Partial Fractions

A function of the form $P(x)/Q(x)$ , where $P(x)$ and $Q(x)$ are polynomials, is called a rational function. Consider the rational function $\frac{x + 7}{(2x - 3)(3x + 4)} = \frac{1}{2x - 3} - \frac{1}{3x + 4}$

The two fractions on the RHS are called partial fractions. To integrate the rational function on the LHS, it is enough to integrate the two fractions on the RHS, which are easily integrable. This is known as method of partial fractions. In case the degree of $P(x)$ (numerator) is not less than that of $Q(x)$ (denominator), we carry out the division of $P(x)$ by $Q(x)$ and reduce the degree of the numerator.

In order to write $P(x)/Q(x)$ in partial fractions, first of all we write

$Q(x) = (x - a)^{k}\ldots\left( x^{2} + \alpha x + \beta \right)^{r}\ldots$ where binomials are different, and then set

$\frac{P(x)}{Q(x)} = \frac{A_{1}}{(x - a)} + \frac{A_{2}}{(x - a)^{2}} + \ldots + \frac{A_{k}}{(x - a)^{k}} + \frac{M_{1}x + N_{1}}{x^{2} + \alpha x + \beta} + \frac{M_{2}x + N_{2}}{\left( x^{2} + \alpha x + \beta \right)^{2}} + \ldots + \frac{M_{r}x + N_{r}}{\left( x^{2} + \alpha x + \beta \right)^{r}} + \ldots$

where $A_{1},\ A_{2},\ \ldots,\ A_{k},\ M_{1},\ M_{2},\ \ldots\ldots,\ M_{r},\ N_{1},\ N_{2},\ \ldots\ldots,\ N_{r}$ are real constants to be determined. These are determined by reducing both sides of the above identity to integral form and equating the coefficients of equal powers of x, which gives a system of linear equations in the coefficient. (This method is called the method of comparison of coefficients). The constants can also be obtained by substituting suitably chosen numerical values of x in both sides of the identity.

Note: Before proceeding to write a rational function as a sum of partial fractions, we should be taken to ascertain that it is either a proper rational fraction or is rewritten as one.

A rational function $\frac{P(x)}{Q(x)}$ is proper if the degree of polynomial $Q(x)$ is greater than the degree of the polynomial $P(x)$ . In case the degree of $P(x)$ greater than or equal to the degree of $Q(x)$ , we first write $\frac{P(x)}{Q(x)} = h(x) + \frac{p(x)}{Q(x)}$ , where $h(x)$ is a polynomial and $p(x)$ is a polynomial of degree less than the degree of polynomial $Q(x)$ .

Case I: Expression $= \frac{1}{a^{2} - x^{2}} = \frac{A}{a + x} + \frac{B}{a - x}$

Let this be equal to $\int\frac{1 + \tan^{2}x}{1 + \cot^{2}x}dx$ , where $A,\ B,\ C$ are constants.

How do we obtain value of $\mathbf{A},\ \mathbf{B}$ and $\mathbf{C}$ ? A, put $x - 1 = 0$ i.e. $x = 1$ in the fraction except in the factor $x - 1$ .

$\therefore\frac{1}{a^{2} - x^{2}} = \frac{A}{a + x} + \frac{B}{a - x} = A = \frac{1}{15}$

Similarly for getting B, put $x + 2 = 0$ i.e. $x = - 2$ except in $x + 2$ .

$\therefore\ \frac{1}{a^{2} - x^{2}} = \frac{A}{a + x} + \frac{B}{a - x} = B = \frac{1}{3}$

For getting $C$ put $2x + 3 = 0$ i.e. $x = 3/2$ in the given fraction except in the factor $2x + 3$

$\begin{matrix} & \ \therefore\ \frac{1}{\left( 1 - \frac{3}{2} + 1 \right)\left( - \frac{3}{2} + 2 \right)} = C = \frac{1}{\frac{- 5}{4}} = - \frac{4}{5} = B = \frac{1}{3} \\ & \text{~Hence~}\frac{1}{a^{2} - x^{2}} = \frac{A}{a + x} + \frac{B}{a - x} = \frac{1}{15(x - 1)} + \frac{1}{3(x + 2)} + \frac{4}{5(2x + 3)} \end{matrix}$

Case II: Expression $= \frac{3x^{2} + 2x + 1}{(x + 1)(x - 1)(2x + 1)}$

This is a proper fraction.

∴ deg (numr) < deg (dear)

So it must be expressed as an algebraic sum of lower degree partial fractions.

Let $\frac{3x^{2} + 2x + 1}{(x + 1)(x - 1)(2x + 1)} = \frac{A}{(x - 1)} + \frac{B}{(x - 2)} + \frac{C}{(2x + 1)}$

For proper fraction the process is same as that for linear non-repeated factors. To find A, put $x + 1 = 0$ i.e. $x = - 1$ in the given fraction except in the term $x + 1$ .

$\therefore\frac{3 - 2 + 1}{( - 1 - 2)( - 2 + 1)} = A = \frac{2}{3}$

To find A, put $A,\ x + 1 = 0$ i.e. $x = 2$ in the given fraction except in the term x 22

$\therefore\frac{{3.2}^{2} + 2.2 + 1}{(2 + 1)(2.2 + 1)} = B = \frac{17}{15}$

To find $C$ , put $2x + 1 = 0$ i.e. $x = - 1/2$ in the fraction except in the factor $2x + 1$

$\therefore\frac{3\left( \frac{1}{4} \right) + 2\left( - \frac{1}{2} \right) + 1}{\left( - \frac{1}{2} + 1 \right)\left( - \frac{1}{2} - 2 \right)} = C = - \frac{3}{5}$

Hence $\frac{3x^{2} + 2x + 1}{(x + 1)(x - 1)(2x + 1)} = \frac{2}{3(x + 1)} + \frac{17}{15(x - 2)} - \frac{3}{5(2x + 1)}$

Case III: Expression $= \frac{3x^{2} + 4x + 5}{2x^{2} - 3x + 1}$

This is not a proper fraction, deg (numerator) = deg (denominator). It can be expressed as the sum of an integral polynomial and a polynomial fraction by simple division process.

Numerator $= 3x^{2} + 4x + 5 = \frac{3}{2}\left( 2x^{2} - 3x + 1 \right) + \frac{17}{2}x + \frac{7}{2}$

Hence the given expression $= \frac{3}{2} + \frac{17x + 7}{2\left( 2x^{2} - 3x + 1 \right)} = \frac{3}{2} + \frac{17x + 7}{2(2x - 1)(x - 1)}$

Let $\frac{17x + 7}{(2x + 1)(x + 1)} = \frac{A}{2x - 1} + \frac{B}{x - 1}$

$\therefore\ A = \frac{\frac{17}{2} + 7}{\frac{1}{2} - 1} = - 31$ and $B = \frac{17 + 7}{1} = 24$

Therefore the expression $= \frac{3x^{2} + 4x + 5}{2x^{2} - 3x + 1} = \frac{3}{2} - \frac{31}{2(2x - 1)} + \frac{12}{x - 1}$

Case IV: Expression $= \frac{3x^{3} + 2x^{2} + x + 1}{x^{2} + 3x + 2}$

This is not a proper fraction. Hence by division process it is to be expressed as the sum of an integral polynomial and a fraction.

Now $3x^{3} + 2x^{2} + x + 1 = 3x\left( x^{2} + 3x + 2 \right) - 7\left( x^{2} + 3x + 2 \right) + (16x + 15)$

Hence given polynomial $\frac{1}{2}$

Now the $2^{\text{nd~}}$ term is a fraction hence it can be expressed as a sum of partial fraction

$\frac{1}{2}ln\left| \frac{2 + sinx - cosx}{2 - sinx + cosx} \right| - \frac{1}{2\sqrt{2}}\tan^{- 1}\left( \frac{sinx + cosx}{\sqrt{2}} \right)$

To find A put $x + 1 = 0$ i.e. $x = - 1$ in the fraction except in the factor $x + 1$

$\therefore\frac{16( - 1) + 15}{( - 1 + 2)} = A + - 1$

To find $\mathbf{B}$ put $x + 2 = 0$ i.e. $x = - 2$ in the fraction except in the factor $x + 2$ .

$\therefore\frac{16( - 1) + 15}{( - 1 + 2)} = B = 17$

∴ The given expression $= 3x - 7 - \frac{1}{x + 1} + \frac{17}{x + 2}$

Case V: Expression $= \frac{x + 5}{(x - 2)^{2}}$ has a repeated (twice) linear factors in the denominator.

Let $\frac{x + 5}{(x - 2)^{2}} = \frac{A}{x - 2} + \frac{B}{(x - 2)^{2}}\ \therefore x + 5 = A(x - 2) + B$

Comparing like terms:

$A = 1; - 2\text{ }A + B = 5$ or $B = 5 + 2 = 7$

$\therefore\frac{x + 5}{(x - 2)^{2}} = - \frac{1}{x - 2} + \frac{7}{(x - 2)^{2}}$

Case VI: Expression $= \frac{2x + 7}{(x + 1)\left( x^{2} + 4 \right)}$ has a quadratic factor in the denominator.

Let $\frac{2x + 7}{(x + 1)\left( x^{2} + 4 \right)} = \frac{A}{x + 1} + \frac{Bx + C}{x^{2} + 4}$

$\therefore 2x + 7 = A\left( x^{2} + 4 \right) + (Bx + C) + (x + 1)$

put $x = - 1\ \therefore 5 = 5\text{ }A$ or $A = 1$

Comparing the like terms:

$\begin{matrix} & 0 = A + B \Rightarrow B = - A = - 1 \\ & 7 = 4A + C\ \therefore C = 7 - 4A = 3 \\ & \ \therefore\frac{2x + 7}{(x + 1)\left( x^{2} + 4 \right)} = \frac{1}{x + 1} + \frac{( - x + 3)}{x^{2} + 4} \end{matrix}$

Or

To obtain value of $A,\ B,\ C$ from $2x + 7 = A\left( x^{2} + 4 \right) + (Bx + C)(x + 1)$

i.e. $2x + 7 = (A + B)x^{2} + (Bx + C)x + 4A + C$

Equating the coefficients of identical powers of $x$ , we get

$A + B = 0,\ B + C = 2$ and $4A + C = 7$ . Solving we get $A = 1,\ B = - 1,\ C = 3$ .

Case VII: Find the partial fractions of $\frac{2x + 1}{(3x + 2)\left( 4x^{2} + 5x + 6 \right)}$

Let $\frac{2x + 1}{(3x + 2)\left( 4x^{2} + 5x + 6 \right)} = \frac{A}{3x + 2} + \frac{Bx + C}{4x^{2} + 5x + 6}$ , then

$2x + 1 = A\left( 4x^{2} + 5x + 6 \right) + (3x + 2)(Bx + C)$ where $A,\ B$ , and $C$ are constants.

For B and C, compare coefficients of $x^{2}$ and the constant terms both sides i.e. $4\text{ }A + 3\text{ }B = 0$

$\therefore B = - \frac{4}{3}A$ and $6A + 2C = 1,\ 2C = 1 - 6A$

Substituting for A, B, C, the integral can be easily solved.

Case VIII: Find the partial fraction of $\frac{2x^{4} + 2x^{2} + x + 1}{x\left( x^{2} + 1 \right)}$

$\begin{matrix} & \frac{2x^{4} + 2x^{2} + x + 1}{x\left( x^{2} + 1 \right)} = \frac{A}{x} + \frac{Bx + C}{\left( x^{2} + 1 \right)^{2}} + \frac{Ex + F}{\left( x^{2} + 1 \right)} \\ & 2x^{4} + 2x^{2} + x + 1 = A\left( x^{2} + 1 \right)^{2} + (Bx + C)x + (Bx + F)x\left( x^{2} + 1 \right) \end{matrix}$

Comparing the coefficients of $x^{4},\ x^{3},\ x^{2},\ x$ and constant term

$\therefore A + E = 2,\ F = 0,\ 2A + B + E = 0,\ C = 1,\ A = 1$

∴ We get $A = 1,\ B = - 1,\ C = 1,\ E = 1,\ F = 0$

Hence the partial fraction is $\frac{1}{x} + \frac{1 - x}{\left( 1 + x^{2} \right)^{2}} + \frac{x}{1 + x^{2}}$

Illustration 17:

Evaluate $\int\frac{cosx}{(1 + sinx)(2 + sinx)}dx$ .

Solution:

Put $sinx = t \Rightarrow cosxdx = dt$

$\begin{matrix} & \ \int_{}^{}\ \frac{cosxdx}{(1 + sinx)(2 + sinx)} = \int_{}^{}\ \frac{dt}{(1 + t)(2 + t)} = \int_{}^{}\ \left\lbrack \frac{1}{t + 1} - \frac{1}{t + 2} \right\rbrack dt \\ & \ = log(t + 1) - log(t + 2) + c \\ & \ = log\left\lbrack \frac{t + 1}{t + 2} \right\rbrack + c = log\left\lbrack \frac{1 + sinx}{2 + sinx} \right\rbrack + c \end{matrix}$

Illustration 18:

Evaluate $\int_{}^{}\ \frac{\mathbf{3x + 1}}{(\mathbf{x} - \mathbf{2})^{\mathbf{2}}(\mathbf{x} + \mathbf{2})}$ .

Solution:

We have, $\frac{3x + 1}{(x - 2)^{2}(x + 2)} = \frac{A}{(x - 2)} + \frac{B}{(x - 2)^{2}} + \frac{C}{(x + 2)}$

$3x + 1 = A(x - 2)(x + 2) + B(x + 2) + C(x - 2)^{2}\#(i)$

Putting $x = 2$ and-2 successively in equation (i), we get $B = \frac{7}{4},\ C = \frac{- 5}{16}$

Now, we put $x = 0$ and get $A = \frac{5}{16}$

$\int\frac{3x + 1}{(x - 2)^{2}(x + 2)}dx = \frac{5}{16}\int\frac{dx}{x - 2} + \frac{7}{4}\int\frac{dx}{(x - 2)^{2}}\frac{- 5}{16}\int\frac{dx}{x + 2}$

$= \frac{5}{16}log(x - 2) - \frac{7}{4}\frac{1}{(x - 2)} - \frac{5}{16}log(x + 2) + c$

$= \frac{5}{16}log\left| \frac{x - 2}{x + 2} \right| - \frac{7}{4(x - 2)} + c$

Illustration 19:

Evaluate $\int\frac{xdx}{x^{3} + 1}$ .

Solution:

Since $x^{3} + 1 = (x + 1)\left( x^{2} - x + 1 \right)$ (the second factor is not a product of linear factors), the partial fractions of the given integer will have the form.

$\frac{x}{x^{3} + 1} = \frac{A}{x + 1} + \frac{Bx + D}{x^{2} - x + 1}$

Hence, $x = A\left( x^{2} - x + 1 \right) + (Bx + D)(x + 1) = (A + B)x^{2} + ( - A + B + D)x + (A + D)$ .

Equating the coefficients of equal powers of $x$ , we get $A = - 1/3,\ B = 1/3,\ D = 1/3$ .

$\text{~Thus~}I = - \frac{1}{3}\int_{}^{}\ \frac{dx}{x + 1} + \frac{1}{3}\int_{}^{}\ \frac{x + 1}{x^{2} - x + 1}dx = - \frac{1}{3}ln|x + 1| + \frac{1}{3}l_{1}$

To calculate the integral $I_{1} = \int_{}^{}\ \frac{x + 1}{x^{2} - x + 1}dx$ , express $(x + 1) = I\left( \right.\$ d.c. of $\left. \ x^{2} - x + 1 \right) + m \Rightarrow (x + 1) = I(2x - 1) + m = 2xI - I + m$

Comparing the coefficients of like powers of $x$ , we get $I = 1/2$ and $m = 3/2$

$\begin{matrix} \Rightarrow l_{1} & \ = \frac{1}{2}\int_{}^{}\ \frac{2x - 1}{x^{2} - x + 1}dx + \frac{3}{2}\int_{}^{}\ \frac{dx}{x^{2} - x + 1} \\ & \ = \frac{1}{2}ln\left| x^{2} - x + 1 \right| + \frac{3}{2}\left( \frac{2}{\sqrt{3}} \right)\tan^{- 1}\frac{x - 1/2}{\sqrt{3}/2} + c_{1} \\ & \ = \frac{1}{2}ln\left| x^{2} - x + 1 \right| + \sqrt{3}\tan^{- 1}\left( \frac{2x - 1}{\sqrt{3}} \right) + c_{1} \end{matrix}$

Hence $I = - \frac{1}{3}ln|x + 1| + \frac{1}{6}ln\left| x^{2} - x + 1 \right| + \frac{1}{\sqrt{3}}\tan^{- 1}\left( \frac{2x - 1}{\sqrt{3}} \right) + c$

Indefinite Integration

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