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  • Need solution for rd sharma maths class 12 chapter Indefinite Integrals exercise 18.2 question 29

Need solution for rd sharma maths class 12 chapter Indefinite Integrals exercise 18.2 question 29

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-cot x- cosec x +c

Hint: To solve this equation we will add 1+cos x

\begin{aligned} &\text { Given: } \int \frac{1}{1-\cos x} d x\\ &\text { Solution: } \int \frac{1}{1-\cos x} d x\\ &\text { Multiplying } 1+\cos x \text { to numerator and denominator }\\ &=\int \frac{1+\cos x}{(1-\cos x)(1+\cos x)} d x\\ &=\int \frac{1+\cos x}{\left(1-\cos ^{2} x\right)} d x \quad\left[\begin{array}{l} (a-b)(a+b)=a^{2}-b^{2} \\ 1-\cos ^{2} x=\sin ^{2} x \end{array}\right]\\ &=\int \frac{1+\cos x}{\sin ^{2} x} d x\\ &=\int \frac{1}{\sin ^{2} x} d x+\int \frac{\cos x}{\sin ^{2} x} d x\\ &=\int \cos e c^{2} x d x+\int \cot x \cos e c x d x\\ &=-\cot x+(-\cos e c x)+c\\ &I=-\cot x-\cos e c x+c \end{aligned}

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