# 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 .