Methods of Integration

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, generally three rules of integration are used.

(1) Integration by substitution or by change of the independent variable.

(2) Integration by parts.

(3) Integration by partial fractions.

INTEGRATION BY SUBSTITUTION

Direct Substitution

If integral is of the form $\int$ f(g(x)) g'(x) dx, then put g(x) = t, provided $\int \text{f(t) dt}$ exists.

(i). $\int \frac{f^{\prime }(x)}{f(x)}dx$ = ln |f (x)| + c

Put f (x) = t $\Rightarrow$ f' (x) dx = dt $\Rightarrow$ $\int \frac{dt}{t}$ = ln |t| + c = ln |f (x)| + c.

(ii). $\int \frac{f^{\prime }(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+c$

Put f (x) = t

$\int \frac{dt}{\sqrt{t}}=2\sqrt{t}+c$ = $2\sqrt{f(x)}+c$.

Illustration -1: Evaluate $\int \frac{\sin \left(\ln x\right)}{x}dx$.

Solution: Let lnx = t. Then dt = $\frac{1}{x}$dx

Hence I = òsint dt = -cost + c = -cos(lnx) + c

Illustration -2: Evaluate $\int \frac{x^3{\tan }^{-1}{x}^4}{1+x^8}dx$.

Solution: Put tan^–1 x⁴ = t $\Rightarrow$ $\frac{4x^3}{1+x^8}dx=dt$

I = $\frac{1}{4}\int tdt$ = $\frac{1}{4}\frac{t^2}{2}+A$ = $\frac{1}{8}$(tan^–1 x⁴)² + A.

Standard Substitutions:

$\bullet$For terms of the form x² + a² or $\sqrt{x^2+a^2}$, put x = a tan$\theta$ or a cotq

For terms of the form x² - a² or $\sqrt{x^2-a^2}$, put x = a sec $\theta$ or a cosecq

(A) For terms of the form a² - x² or $\sqrt{a^2-x^2}$, put x = a sin $\theta$ or a cosq

$\bullet$If both $\sqrt{a+x}$, $\sqrt{a-x}$ are present, then put x = a cos $\theta$

$\bullet$For the type $\sqrt{\left(x-a\right)\left(b-x\right)}$, put x = a cos²$\theta$+ b sin²$\theta$

$\bullet$For the type${\left(\sqrt{x^2+a^2}\pm x\right)}^n\text{ or }{\left(\text{x}\pm \sqrt{{\text{x}}^{\text{2}}-a^2}\right)}^n$, put the expression within the

bracket = t.

$\bullet$For the type ${\left(x+a\right)}^{-1-\frac{1}{n}}{\left(x+b\right)}^{-1+\frac{1}{n}}or{\left(\frac{x+b}{x+a}\right)}^{\frac{1}{n}-1}\frac{1}{{\left(x+a\right)}^2}$ (n $\in$N, n> 1), put $\frac{x+b}{x+a}=t$.

$\bullet$For$\frac{1}{{\left(x+a\right)}^{n_1}{\left(x+b\right)}^{n_2}}$, n₁,n₂ $\in$N (and > 1), again put (x + a) = t (x + b)

Illustration -3: Evaluate $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$.

Solution: $\int \frac{1-\sqrt{x}}{\sqrt{1-x}}dx=\int \frac{1}{\sqrt{1-x}}dx-\int \sqrt{\frac{x}{1-x}}dx=-2\sqrt{1-x}-\int \sqrt{\frac{x}{1-x}}dx$ = I₂

Put 1 –x = t²

dx = 2t dt

I = –$\sqrt{1-x}+\sqrt{x}\sqrt{1-x}+{\sec }^{-1}\sqrt{1-x}+k$

Derived Substitution:

Some time it is useful to write the integral as a sum of two related integrals which can be evaluated by making suitable substitutions.

Illustrations of such integrals are:

A. Algebraic Twins

$\int \frac{\text{2}{\text{x}}^{\text{2}}}{{\text{x}}^{\text{4}}+1}dx=\int \frac{x^2+1}{x^4+1}dx+\int \frac{x^2-1}{x^4+1}dx$,

$\int \frac{\text{2}}{{\text{x}}^{\text{4}}+1}dx=\int \frac{x^2+1}{x^4+1}dx-\int \frac{x^2-1}{x^4+1}dx$,

$\int \frac{2x^2}{\left(x^4+1+k{x}^2\right)}dx,\int \frac{2}{\left(x^4+1+k{x}^2\right)}dx$.

B. Trigonometric twins

$\int \sqrt{\tan \text{ x}}dx,\int \sqrt{\text{cot x}}dx$,

$\int \frac{1}{\left({\sin }^{4}x+{\cos }^{4}x\right)}dx$, $\int \frac{1}{\left({\sin }^{6}x+{\cos }^{6}x\right)}dx$, $\int \frac{\pm \sin x\pm \cos x}{a+b\sin x\cos x}dx$.

Method of evaluating these integral are illustrated by mean of the following Illustrations:

Illustration -5: Evaluate $\int \frac{5}{1+x^4}dx$.

Solution: I = $\int \frac{5}{1+x^4}dx$ = $\frac{5}{2}\int \frac{2}{1+x^4}dx$ = $\frac{5}{2}\left[\int \frac{1+x^2}{1+x^4}+\frac{1-x^2}{1+x^4}dx\right]$

= $\frac{5}{2}\left[\int \frac{x^2+1}{1+x^4}-\frac{x^2-1}{1+x^4}dx\right]$= $\frac{5}{2}\left[\int \left(\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}-\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\right)dx\right]$ = $\frac{5}{2}\left[I_1-I_2\right]$

For I₁, we write x -$\frac{1}{x}$ = t $\Rightarrow$ (1 +$\frac{1}{x^2}$)dx = dt

I₁ = $\int \frac{dt}{t^2+{\left(\sqrt{2}\right)}^2}$

= $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)$ = $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1/x}{\sqrt{2}}\right)$

For I₂ , we write x + $\frac{1}{x}$= t $\Rightarrow$ (1 - $\frac{1}{x^2}$)dx = dt

$\Rightarrow$ I₂= $\int \frac{dt}{t^2-{\left(\sqrt{2}\right)}^2}$ = $\frac{1}{2\sqrt{2}}\ln \left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|$ = $\frac{1}{2\sqrt{2}}\ln \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|$

Combining the two integrals, we get

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