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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 10 Maths Textbook Solution.

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Answer:\frac{4}{2-\sin x}+3 \log |2-\sin x|+c

Hint: Use integration method

Given: \int \frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x

Solution: I=\int \frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x

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

            =\int \frac{(3 \sin x-2) \cos x}{4+\sin ^{2} x-4 \sin x} d x

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

\begin{aligned} &\text { Let } t=2-\sin x \\ &\sin x=2-t \\ &\qquad d t=-\cos x d x \end{aligned}

\begin{aligned} &I=\int \frac{3(2-t)-2}{t^{2}}(-d t) \\ &=-\int \frac{4-3 t}{t^{2}} d t \end{aligned}

\begin{aligned} &=-4 \int \frac{d t}{t^{2}}+3 \int \frac{1}{t} d t \\ &I=-4\left(\frac{-1}{t}\right)+3 \log t+c_{1} \\ &I=\frac{4}{2-\sin x}+3 \log |2-\sin x|+c \end{aligned}

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