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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 10 Maths Textbook Solution.

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Answer: -2 \cot \left(\frac{x}{2}\right)-x+c

Hint: \text { To solve this equation we use half angle formula }

Given: \int \frac{1+\cos x}{1-\cos x} d x

Solution: \int \frac{1+\cos x}{1-\cos x} d x

=\int \frac{1+2 \cos ^{2} \frac{x}{2}-1}{1-1+2 \sin ^{2} \frac{x}{2}} d x                                \text { [By using half angle formula] }

\begin{aligned} &=\int \frac{2 \cos ^{2} \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}} d x \\ &=\int \cot ^{2} \frac{x}{2} d x \end{aligned}

\begin{aligned} &=\int\left(\cos e c^{2} \frac{x}{2}-1\right) d x \\ &=\int \cos e c^{2} \frac{x}{2} d x-\int d x\left[\int \operatorname{cosec}^{2} x d x=-\cot x+c\right] \\ &=-2 \cot \frac{x}{2}-x+c \end{aligned}

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