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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 20 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 20 Maths Textbook Solution.

Answers (1)

Answer:

-e^{x} \cot \frac{x}{2}+C

Given:

\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x

Hint:

You must know about the derivation of cot x and \int e^{x} d x

Explanation:

Let \mathrm{I}=\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x

        =\int e^{x}\left(\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos _{2}^{x}}{2 \sin ^{2} \frac{x}{2}}\right) d x                                                      \left[\begin{array}{l} \because \sin ^{2} \theta+\cos ^{2} \theta=1 ; \sin 2 \theta=2 \sin \theta \cos \theta ; \\ 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2} \end{array}\right]

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

             =\frac{1}{2} \int e^{x}\left(\operatorname{cosec}^{2} \frac{x}{2}-2 \cot \frac{x}{2}\right) d x                                                                                           \left[\because 1+\cot ^{2} \theta=\cos e^{2} \theta\right]

\begin{aligned} &=\frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x-\int \cot \frac{x}{2} e^{x} d x \\ =&-\int \cot \frac{x}{2} \cdot e^{x} d x+\int \cos e c^{2} \frac{x}{2} e^{x} \cdot \frac{1}{2} d x \\ =&-\left[\cot \frac{x}{2} \cdot e^{x}-\int-\operatorname{cosec}^{2} \frac{x}{2} \cdot \frac{1}{2} \cdot e^{x} d x\right]+\frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x \ldots . .\left[\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right] \\ =&-\cot \frac{x}{2} \cdot e^{x}-\frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x+\frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x \\ =&-\cot \frac{x}{2} \cdot e^{x}+C \end{aligned}

      

Posted by

infoexpert21

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