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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 7 Maths Textbook Solution.

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Answer:  \frac{1}{5} \log \left|\frac{\tan x-2}{2 \tan x+1}\right|+c

Given:  \int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x

Hint: Apply substitution method and divide Numerator and Denominator by  \cos ^{2} x


    \int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x

    =\int \frac{1}{2 \sin ^{2} x+\sin x \cos x-4 \sin x \cos x-2 \cos ^{2} x} d x

    =\int \frac{1}{2 \sin ^{2} x-3 \sin x \cos x-2 \cos ^{2} x} d x

On dividing Numerator and Denominator by \cos ^{2} x

=\int \frac{\sec ^{2} x}{2 \tan ^{2} x-3 \tan x-2} d x


\tan x=t                                                                    (Differentiate w.r.t x)

\sec ^{2} d x=d t

Now,\int \frac{1}{2 t^{2}-3 t-2} d t

=\frac{1}{2} \int \frac{1}{t^{2}-\frac{3}{2} t-1} d t

=\frac{1}{2} \int \frac{1}{t^{2}-2 t\left(\frac{3}{4}\right)+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}-1} d t

=\frac{1}{2} \int \frac{1}{\left(t-\frac{3}{4}\right)^{2}-\left(\frac{5}{4}\right)^{2}} d t

=\frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \log \left|\frac{t-\frac{3}{4}-\frac{5}{4}}{t-\frac{3}{4}+\frac{5}{4}}\right|+c                                                                \left[\int \frac{d t}{t^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{t-a}{t+a}\right|+c\right]

\begin{aligned} &=\frac{1}{2} \times \frac{2}{5} \log \left|\frac{2t-4}{2 t+1}\right|+c \\ &=\frac{1}{5} \log \left|\frac{2\tan x-4}{2 \tan x+1}\right|+c \end{aligned}


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