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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 50 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 50 Maths Textbook Solution.

Answers (1)

Answer:e\frac{\pi }{2}

Hint: Use indefinite integral formula then put the limits to solve this integral

Given:

\int_{\frac{\pi }{2}}^{\pi }e^{x}\left ( \frac{1-\sin x}{1-\cos x} \right )dx

Sol:

\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x=\int_{\frac{\pi}{2}}^{\pi} e^{x} \frac{\left(1-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}\right)}{2 \sin ^{2} \frac{x}{2}} d x

                                                                                             \left[\begin{array}{l}\sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos 2 \theta=2 \sin ^{2} \theta\end{array}\right]                    

    =\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1}{2 \sin ^{2} \frac{x}{2}}-\frac{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}}\right) d x

                                                                                              \left[\frac{1}{\sin \theta}=\operatorname{cosec} \theta, \cot \theta=\frac{\sin \theta}{\cos \theta}\right]

=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{\cos e c^{2} \frac{x}{2}}{2}-\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) d x

=\int_{\frac{\pi}{2}}^{\pi} \frac{e^{x} \cos e c^{2} \frac{x}{2}}{2}-\int_{\frac{\pi}{2}}^{\pi} e^{x} \cot \frac{x}{2} d x

=\frac{1}{2}\left[\int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2 \int_{\frac{\pi}{2}}^{\pi} e^{x} \cot \frac{x}{2} d x\right]

=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{x}{2} \int e^{x} d x\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\left(\frac{d\left(\cot \frac{x}{2}\right)}{d x} \int e^{x} d x\right) d x\right\}

Using integration by parts in second term,

=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{x}{2} e^{x}\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\left(-\cos e c^{2} \frac{x}{2}\right) \frac{1}{2} e^{x} d x\right\}

\left[\frac{d}{d x} \cot a x=-\cos e c^{2} a x \cdot a, \int e^{x} d x=e^{x}\right]

=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{\pi}{2} \cdot e^{\pi}-\cot \frac{\pi}{4} e^{\frac{\pi}{2}}\right]+\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \cos e c^{2} \frac{x}{2} e^{x} d x\right\}

=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-1\left(0 . e^{\pi}-1 \cdot e^{\frac{\pi}{2}}\right)-\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \cos e c^{2} \frac{x}{2} e^{x} d x

\left[\cot \frac{\pi}{2}=0, \cot \frac{\pi}{4}=1\right]

=-\left(0-e^{\frac{\pi}{2}}\right)

=e^{\frac{\pi}{2}}

Posted by

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