Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x+a)^n}dx=\frac{1}{(1-n)(x+a)^{n+1}}+C\)

Calculation:​

To solve: \(\rm \int \frac{sec^2\theta}{(tan\theta)(tan\theta-1)(tan\theta-2)}d\theta\)

let tan θ = x  ⇒ sec²θ dθ = dx

⇒ \(\rm \int \frac{sec^2\theta}{(tan\theta)(tan\theta-1)(tan\theta-2)}d\theta = \rm \int \frac{1}{(x)(x-1)(x-2)}dx\)

This integrand is a proper rational fraction. So, by using the form of partial fraction, we can write it as

⇒ \(\rm \frac{1}{(x)(x-1)(x-2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}\)

⇒ 1 = A(x - 1)(x - 2) + B(x)(x - 2) +C(x)(x - 1)

⇒ 1 = (A + B +C)x2 + (-3A - 2B - C)x + 2A

Comparing coefficient of x2, x and constant terms on both sides, we get A + B +C = 0, -3A - 2B - C = 0 and 2A = 1

By solving these equation, we get  A = 1/2,  B = -1 and C = 1/2

⇒ \(\rm \frac{1}{(x)(x-1)(x-2)}=\frac{1/2}{x}-\frac{1}{x-1}+\frac{1/2}{x-2}\)  

⇒ \(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=\frac{1}{2}\int\frac{1}{x}dx-\int\frac{1}{x-1}dx+\frac{1}{2}\int\frac{1}{x-2}dx\)

⇒ \(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=\frac{1}{2}log\left | x \right |-log\left | x-1 \right |+\frac{1}{2}log\left | x-2 \right |+C\)   

⇒ \(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=\frac{1}{2}log\left | x(x-2) \right |-log\left | x-1 \right |+C\)

⇒ \(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=log\sqrt{|x(x-2)|}-log\left | x-1 \right |+C\)

⇒\(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=log|\frac{\sqrt{|x(x-2)|}}{ x-1}|+C\)

Now put x = tanθ in the above equation we get

⇒ \(\rm \int\frac{sec^2\theta}{(tan\theta)(tan\theta-1)(tan\theta-2)}d\theta=log|\frac{\sqrt{|tan\theta(tan\theta-2)|}}{ tan\theta-1}|+C\)