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  • Explain solution rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 42

Explain solution rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 42

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\begin{aligned} & x-\tan \frac{x}{2}+c \\ &\text { Hint: } \mathrm{U} \operatorname{sing} \int \sec ^{2} x d x \\ &\text { Given: } \int \frac{\cos x}{1+\cos x} d x \\ &\text { Solution: } I=\int \frac{\cos x}{1+\cos x} d x \end{aligned}

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

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