Question

Integrate $\int \frac{1}{\left[\left(\sin x-2\cos x\right)\left(2\sin x+\cos x\right)\right]}dx$

Answer 1

Step:1 Simplify the given integral

Given: $\int \frac{1}{\left[\left(\sin x-2\cos x\right)\left(2\sin x+\cos x\right)\right]}dx$

We can simplify it as,

$=\int \frac{1}{2s{\mathrm{in}}^{2}x+\sin x\cos x-4\sin x\cos x-2{\cos }^{2}x}dx$

Dividing the numerator and denominator by ${\cos }^{2}x$,

Now,

$=\int \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\cos }^{2}x$}\right.}}{{\displaystyle \frac{2{\sin }^{2}x-3\sin x\cos x-2{\cos }^{2}x}{{\cos }^{2}x}}}dx$

$=\int \frac{se{c}^{2}x}{2{\tan }^{2}x-3\tan x-2}dx$

Step: 2 Using the method of substitution to find the integral

Put $\tan x=t$ so that $se{c}^{2}xdx=dt$

Now,

$\int \frac{se{c}^{2}x}{2{\tan }^{2}x-3\tan x-2}dx$

$$\begin{gathered} =\int \frac{dt}{2t^2-3t-2} \\ =\frac{1}{2}\int \frac{1}{\left(t^2-{\displaystyle \frac{3t}{4}}-1\right)}dt \\ =\frac{1}{2}\int \frac{1}{{\left(t-{\displaystyle \frac{3}{4}}\right)}^2-{\left({\displaystyle \frac{5}{2}}\right)}^2}dt \end{gathered}$$

Step: 3 Using the appropriate formula to find the value of integral

We know that,

$\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C$

Now,

$$\begin{gathered} =\frac{1}{2}\times \frac{1}{2\left({\displaystyle \frac{5}{4}}\right)}\log \left|\frac{t-{\displaystyle \frac{3}{4}}-{\displaystyle \frac{5}{4}}}{t-{\displaystyle \frac{3}{4}}+{\displaystyle \frac{5}{4}}}\right|+C \\ =\frac{1}{5}\log \left|\frac{t-2}{t+{\displaystyle \frac{1}{2}}}\right|+C \\ =\frac{1}{5}\log \left|\frac{2\tan x-4}{2\tan x+1}\right|+C \end{gathered}$$

Hence, Integral $\int \frac{1}{\left[\left(\sin x-2\cos x\right)\left(2\sin x+\cos x\right)\right]}dx$ is$\frac{1}{5}\log \left|\frac{2\tan x-4}{2\tan x+1}\right|+C$, where $C$ is an arbitrary constant.