Question

Evaluate ∫ 1/((sinx-2cosx)(2sinx+cosx))dx

\(\int\frac{1}{(sinx-2cosx)(2sinx+cosx)}\) dx

Answer 1

Given,

∫ 1/((sinx-2cosx)(2sinx+cosx))dx

Let tanx = t

\(\frac{d}{dx}\) (tanx) = dt

sec²x dx = dt

Now,

\(\int\frac{dt}{2t^2-3t-2}\)

= \(\int\frac{dt}{(2t+1)(t-2)}\)

Now,

\(\frac{1}{(2t+1)(t-2)}\) ≅ \(\frac{A}{2t+1}\) + \(\frac{B}{t-2}\)

1≅ A(t-2)+B(2t+1) 

Equating ‘t’ coeff: - 

0 = A + 2B 

A = - 2B 

Equating constant: - 

1 = - 2A + B 

1 = - 2(- 2B) + B 

1 = 5B

B = \(\frac{1}{5}\) 

A = \(\frac{-2}{5}\)