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

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Answer: 2\left[\tan \frac{x}{2}+\sec \frac{x}{2}\right]+c

Hint: \text { We will add } 1+\sin \frac{x}{2} \text { to the equation }

Given: \int \frac{1}{1-\sin \frac{x}{2}} d x

Solution: \int \frac{1}{1-\sin \frac{x}{2}} d x

\text { Multiplying } 1+\sin \frac{x}{2} \text { in numerator and denominator }

=\int \frac{1+\sin \frac{x}{2}}{\left(1-\sin \frac{x}{2}\right)\left(1-\sin \frac{x}{2}\right)} d x

\begin{aligned} &=\int \frac{1+\sin \frac{x}{2}}{1-\sin ^{2} \frac{x}{2}} d x \\ &=\int \frac{1+\sin \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x \end{aligned}

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

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

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



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