Integration by 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.

Examples of such integrals are:

A. Algebraic Twins

$ \displaystyle \int \frac{2x^2}{x^4 + 1}dx = \int \frac{x^2 + 1}{x^4 + 1}dx + \int \frac{x^2 – 1}{x^4 + 1}dx $

$ \displaystyle \int \frac{2}{x^4 + 1}dx = \int \frac{x^2 + 1}{x^4 – 1}dx + \int \frac{x^2 – 1}{x^4 + 1}dx $

$ \displaystyle \int \frac{2x^2}{x^4 + 1 + kx^2}dx \; , \int \frac{2}{x^4 + 1 + kx^2}dx $

B. Trigonometric twins

$ \displaystyle \int \sqrt{tanx}dx \; , \int \sqrt{cotx}dx $

$ \displaystyle \int \frac{1}{sin^4 x + cos^4 x}dx $

$ \displaystyle \int \frac{1}{sin^6 x + cos^6 x}dx $

$ \displaystyle \int \frac{\pm sinx \pm cosx}{a + bsinx cosx}dx $

Illustration : Evaluate

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

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

(iii) $ \displaystyle \int \sqrt{tanx}dx $

Solution: (i)

$ \displaystyle I = \int \frac{5}{x^4 + 1}dx $

$ \displaystyle = \frac{5}{2} \int \frac{2}{x^4 + 1}dx $

$ \displaystyle = \frac{5}{2} [\int \frac{x^2 +1}{x^4 + 1}dx – \int \frac{x^2 -1}{x^4 + 1} dx] $

Dividing by x2

$ \displaystyle = \frac{5}{2} [\int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}}dx – \int \frac{1 -\frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx] $

$ \displaystyle = \frac{5}{2} [I_1 – I_2 ] $

For I1 we write , $latex \displaystyle x-\frac{1}{x} = t $

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

$ \displaystyle I_1 = \int \frac{dt}{t^2 +(\sqrt{2})^2} $

$ \displaystyle I_1 = \frac{1}{\sqrt{2}}tan^{-1}\frac{t}{\sqrt{2}} $

For I2 we write , $latex \displaystyle x+\frac{1}{x} = t $

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

$ \displaystyle I_2 = \int \frac{dt}{t^2 – (\sqrt{2})^2} $

$ \displaystyle I_2 = \frac{1}{2\sqrt{2}}ln|\frac{t-\sqrt{2}}{t+\sqrt{2}}| $

Combining the two integrals,

$ \displaystyle I = \frac{5}{2} [I_1 – I_2 ] $

(ii) $ \displaystyle I = \int \frac{1}{x^4 + 1 + 5x^2}dx $

$ \displaystyle = \frac{1}{2} \int \frac{2}{x^4 + 1 + 5x^2}dx $

$ \displaystyle = \frac{1}{2} \int \frac{x^2 + 1}{x^4 + 1 + 5x^2}dx – \frac{1}{2} \int \frac{x^2 – 1}{x^4 + 1 + 5x^2}dx $

on dividing by x2

$ \displaystyle = \frac{1}{2} \int \frac{1 + 1/x^2}{x^2 + 1/x^2 + 5 }dx – \frac{1}{2} \int \frac{1 – 1/x^2}{x^2 + 1/x^2 + 5}dx $

$ \displaystyle = \frac{1}{2}(I_1 – I_2) $

For I1 , we write

$ \displaystyle x-\frac{1}{x}=t $

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

$ \displaystyle I_1 = \int \frac{dt}{t^2 + 7} $

$ \displaystyle = \frac{1}{\sqrt{7}}tan^{-1}\frac{t}{\sqrt{7}} $

For I2, we write

$ \displaystyle x +\frac{1}{x} = t $

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

$ \displaystyle I_2 = \int \frac{dt}{t^2 + 3} $

$ \displaystyle = \frac{1}{\sqrt{3}}tan^{-1}\frac{t}{\sqrt{3}} $

Combining the two results,

$ \displaystyle I = \frac{1}{2}(I_1 – I_2) $

(iii) $ \displaystyle I = \int \sqrt{tanx}dx $

Put tanx = t2

=> sec2x dx = 2t dt

$ \displaystyle dx = \frac{2t dt}{sec^2 x} $

$ \displaystyle dx = \frac{2t dt}{1 + tan^2 x} $

$ \displaystyle dx = \frac{2t dt}{1 + t^4} $

$ \displaystyle I = \int t \frac{2t dt}{1 + t^4} $

$ \displaystyle = \int \frac{2t^2 dt}{1 + t^4} $

$ \displaystyle I = \int \frac{t^2 +1}{t^4 + 1}dt + \int \frac{t^2 – 1}{t^4 + 1}dt $

This can be solved by the method used above .

Also Read :

Indefinite Integral : Basic Concepts
Integration by substitution(Direct & Indirect Substitution)
Integration by Derived Substitution
Integration by Parts
Integration by Partial Fractions

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