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NCERT Class XII Mathematics Chapter - - Solutions

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Evaluate the definite integrals in Exercises 253 to 261.
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Question : 253 of 271
Marks: +1, -0
∫π/2πex(1−sin⁡x1−cos⁡x)dx\int\limits_{\pi/2}^{\pi} e^x \left( \frac{1-\sin x}{1-\cos x} \right) dx
Solution:  
Let I=∫π/2πex(1−sin⁡x1−cos⁡x)dxI = \int\limits_{\pi/2}^{\pi} e^x \left( \frac{1-\sin x}{1-\cos x} \right) dx
=∫π/2πex(1−2sin⁡x2cos⁡x22sin⁡2x2)dx= \int\limits_{\pi/2}^{\pi} e^x \left( \frac{1 - 2 \sin\frac{x}{2} \cos\frac{x}{2}}{2 \sin^2\frac{x}{2}} \right) dx
=∫π/2πex(12csc⁡2x2−cot⁡x2)dx= \int\limits_{\pi/2}^{\pi} e^x \left( \frac{1}{2} \csc^2\frac{x}{2} - \cot\frac{x}{2} \right) dx
=−∫π/2πex(cot⁡x2−12csc⁡2x2)dx= -\int\limits_{\pi/2}^{\pi} e^x \left( \cot\frac{x}{2} - \frac{1}{2} \csc^2\frac{x}{2} \right) dx
=−[excot⁡x2]π/2π= -\left[ e^x \cot\frac{x}{2} \right]_{\pi/2}^{\pi}
=−[eπcot⁡π2−eπ/2cot⁡π4]= -\left[ e^{\pi} \cot\frac{\pi}{2} - e^{\pi/2} \cot\frac{\pi}{4} \right]
=−[0−eπ/2⋅1]=eπ/2= -\left[ 0 - e^{\pi/2} \cdot 1 \right] = e^{\pi/2}
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