Integration by Substitution( Direct & Indirect )

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:

There are following types of substitutions

Integration by substitution(Direct Substitution):

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

Illustration : Evaluate

$\displaystyle \int \frac{sin(lnx)}{x}dx$

Solution:

$\displaystyle I = \int \frac{sin(lnx)}{x}dx$

Let lnx = t , then dt = (1/x) dx

Hence $\displaystyle I= \int sint dt$

= −cost + c

= −cos(lnx) + c

Exercise :

Integrate the following functions

(i) $\displaystyle \int e^{tan^{-1}x } \frac{1}{1+x^2}dx$

(ii) $\displaystyle \int \frac{3sinx + 4cosx}{4sinx-3cosx}dx$

(iii)$\displaystyle \int \sqrt[3]{\frac{sin^n x}{cos^{n+6}x}} dx$

(iv) $\displaystyle \int \frac{4x + 6}{x^2 + 3x + 100} dx$

Standard Substitutions:

∎ For terms of the form $\displaystyle x^2 +a^2 \; or , \sqrt{x^2 + a^2}$

put x = a tanθ or a cotθ

∎ For terms of the form $\displaystyle x^2 – a^2 \; or , \sqrt{x^2 – a^2}$

put x = a sec θ or a cosecθ

∎ For terms of the form $\displaystyle a^2 – x^2 \; or , \sqrt{a^2 – x^2}$

put x = a sin θ or a cosθ

∎ the type $\displaystyle (\sqrt{x^2 +a^2 } \pm x)^n \; or ,( x \pm \sqrt{x^2 – a^2})$

put the expression within the bracket = t.

∎ For If both $\displaystyle \sqrt{a+x} \; , \sqrt{a-x}$ are present ,

then put x = a cosθ.

∎ For the type $\displaystyle \sqrt{(x-a)(b-x)}$

put x = a cos2 θ + b sin2 θ

Illustration : Evaluate

(i) $\displaystyle \int \frac{dx}{(x+1)^{6/5}(x-3)^{4/5}}$

(ii) $\displaystyle \int \frac{dx}{\sqrt{(x-a)(b-x)}}$

Solution: (i)

$\displaystyle I = \int \frac{dx}{(x+1)^{6/5}(x-3)^{4/5}}$

$\displaystyle \int \frac{dx}{(x+1)^2 (\frac{x-3}{x+1})^{4/5}}$

Put , $\displaystyle \frac{x-3}{x+1} = t$

$\displaystyle dt = \frac{4}{(x+1)^2}dx$

$\displaystyle I= \frac{1}{4}\int \frac{dt}{t^{4/5}}$

$\displaystyle = \frac{5}{4}t^{1/5} +c$

(ii)    $\displaystyle I = \int \frac{dx}{\sqrt{(x-a)(b-x)}}$

Put x = acos2θ + bsin2θ, the given integral becomes.

$\displaystyle I = \int \frac{2(b-a)sin\theta cos\theta d\theta}{\sqrt{(acos^2 \theta +bsin^2\theta -a)(b-acos^2\theta -bsin^2\theta)}}$

$\displaystyle = \int \frac{2(b-a)sin\theta cos\theta d\theta}{(b-a)sin\theta cos\theta}$

$\displaystyle = 2\int d\theta$

$\displaystyle = 2\theta + c$

$\displaystyle = 2sin^{-1}\sqrt{\frac{x-a}{b-a}} +c$

Integration by substitution (Indirect Substitution):

If the integrand is of the form f(x) g(x) , where g(x) is a function of the integral of f(x) , then put integral of f(x) = t

Illustration : Evaluate

(i) $\displaystyle \int \frac{\sqrt{x}}{\sqrt{x^3 + a^3}} dx$

(ii) $\displaystyle \int \frac{sinx + cosx}{9 + 16sin2x} dx$

Solution: (i)

$\displaystyle I = \int \frac{\sqrt{x}}{\sqrt{x^3 + a^3}} dx$

Integral of the numerator $\displaystyle = \frac{x^{3/2}}{3/2}$

Put x3/2 = t .

We get $\displaystyle I = \frac{2}{3}\int \frac{dt}{\sqrt{t^2 + a^3}}$

$\displaystyle = \frac{2}{3}ln |x^{3/2} + \sqrt{x^3 + a^3} | + c$

(ii) $\displaystyle I = \int \frac{sinx + cosx}{9 + 16sin2x} dx$

$\displaystyle = \int \frac{sinx + cosx}{25-16 + 16sin2x} dx$

$\displaystyle = \int \frac{sinx + cosx}{25-16(sin^2 x + cos^2 x) + 16sin2x} dx$

$\displaystyle = \int \frac{sinx + cosx}{25 – 16 (sinx-cosx)^2 } dx$

Put t = sinx − cosx

=> dt = (cosx + sinx)dx

=> $\displaystyle I = \int \frac{dt}{25 – 16 t^2 }$

$\displaystyle = \frac{1}{40} ln|\frac{5-4t}{5+4t}| +c$