Question

Solve $ \int {\dfrac{1}{{(\sin x - 2\cos x)(2\sin x + \cos x)}}} dx $

Answer 1

Hint: Given question is not possible to integrate directly since it is not in any standard form suitable for integration. So we need to simplify or modify the denominator to get the form which can be integrated. After modifying the denominator, converting the multiple to a single term will make it easy to solve. By using the formulas given below we can solve the given equation.

Formulas Used:
 $ \cos 2x = {\cos ^2}x - {\sin ^2}x $
 $ \sin 2x = 2\sin x\cos x $
 $ \cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} $
 $ \int {\dfrac{1}{{{x^2} - {a^2}}}} dx = \dfrac{1}{{2a}}\log [\dfrac{{x - a}}{{x + a}}] + C $

Complete step-by-step answer:
Let us consider the given question
 $ \int {\dfrac{1}{{(\sin x - 2\cos x)(2\sin x - \cos x)}}} dx $
On multiplying the terms in the denominator we get the following
 $ = \int {\dfrac{1}{{2{{\sin }^2}x + \sin x\cos x - 4\sin x\cos x - 2{{\cos }^2}x}}} dx $
On rearranging the terms in the denominator and adjusting them gives the following
 $ = \int {\dfrac{1}{{2({{\sin }^2}x - {{\cos }^2}x) - \dfrac{3}{2}(2\sin x\cos x)}}dx} $
We know that
 $ {\cos ^2}x - {\sin ^2}x = \cos 2x $ and
 $ 2\sin x\cos x = \sin 2x $
 $ = \int {\dfrac{1}{{ - 2\cos 2x - \dfrac{3}{2}\sin 2x}}dx} $
For easy solving let us assume $ \tan x = t $
On differentiating $ \tan x = t $ on both sides
 $ \dfrac{d}{{dx}}(\tan x) = \dfrac{d}{{dx}}(t) $
 $ {\sec ^2}x = \dfrac{{dt}}{{dx}} $
We know that $ {\sec ^2}x = 1 + {\tan ^2}x = 1 + {t^2} $
 $ 1 + {t^2} = \dfrac{{dt}}{{dx}} $
 $ dx = \dfrac{{dt}}{{1 + {t^2}}} $
 $ \cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \dfrac{{1 - {t^2}}}{{1 + {t^2}}} $
 $ \sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}} = \dfrac{{2t}}{{1 + {t^2}}} $
On substituting the values of $ \cos 2x $ and $ \sin 2x $ in terms of $ t $ and $ dx $ in terms of $ dt $ we get
 $ = - \int {\dfrac{1}{{2(\dfrac{{1 - {t^2}}}{{1 + {t^2}}}) + \dfrac{3}{2}(\dfrac{{2t}}{{1 + {t^2}}})}}} {\text{ }}\dfrac{{dt}}{{(1 + {t^2})}} $
 $ = - \int {\dfrac{{(1 + {t^2})}}{{2(1 - {t^2}) + \dfrac{3}{2}(2t)}}} {\text{ }}\dfrac{{dt}}{{(1 + {t^2})}} $
 $ = - \dfrac{1}{2}\int {\dfrac{1}{{(1 - {t^2}) + \dfrac{3}{2}t}}dt} $
 $ = \dfrac{1}{2}\int {\dfrac{1}{{{t^2} - \dfrac{3}{2}t - 1}}} dt $
Adding $ + \dfrac{9}{{16}} $ and $ - \dfrac{9}{{16}} $ to the denominator will change that into a form that can be integrated.
 $ = \dfrac{1}{2}\int {\dfrac{1}{{{t^2} - \dfrac{3}{2}t - 1 + \dfrac{9}{{16}} - \dfrac{9}{{16}}}}} dt $
The denominator can be written as follows
 $ = \dfrac{1}{2}\int {\dfrac{1}{{{{(t - \dfrac{3}{4})}^2} - {{(\dfrac{5}{4})}^2}}}} dt $
Now it is in the form of $ \int {\dfrac{1}{{{x^2} - {a^2}}}dx} $
We know that $ \int {\dfrac{1}{{{x^2} - {a^2}}}} dx = \dfrac{1}{{2a}}\log [\dfrac{{x - a}}{{x + a}}] + C $
Where $ C $ is the Integration Constant.
Therefore,
 $ = \dfrac{1}{2} \times \dfrac{1}{{(2 \times \dfrac{5}{4})}}\log [\dfrac{{(t - \dfrac{3}{4}) - \dfrac{5}{4}}}{{(t - \dfrac{3}{4}) + \dfrac{5}{4}}}] + C $
Since the boundaries are not known (Indefinite Integration) $ C $ should be added.
 $ = \dfrac{1}{5}\log [\dfrac{{4t - 8}}{{4t + 2}}] + C $
 $ = \dfrac{1}{5}\log [\dfrac{{2t - 4}}{{2t + 1}}] + C $
On replacing $ t $ with $ \tan x $ we get the following result
 $ = \dfrac{1}{5}\log [\dfrac{{2\tan x - 4}}{{2\tan x + 1}}] + C $
Therefore, $ \int {\dfrac{1}{{(\sin x - 2\cos x)(2\sin x + \cos x)}}dx = \dfrac{1}{5}} \log [\dfrac{{2\tan x - 4}}{{2\tan x + 1}}] + C $

Note: In the above solution on simplifying the initial question we got the simplified question in $ \cos 2x $ and $ \sin 2x $ which is not suitable for integration. Converting the two terms into $ \tan x $ can help us. In this type of problems involving $ \sin $ and $ \cos $ together and not in suitable form for integration it is better to convert them into $ \tan $ . In order to convert $ \cos 2x $ and $ \sin 2x $ to $ \tan x $ we need to divide and multiply the equation with $ {\cos ^2}x $ .