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  • Evaluate the definite integrals in Exercises 25 to 33. 25

Evaluate the definite integrals in Exercises 25 to 33.

    Q25.    \int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx

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

Since, we have  e^x multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,

 \int e^x(f(x)+f'(x))dx=e^xf(x)

So,

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

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

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

=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2}cosec^2\frac{x}{2}-cot\frac{x}{2}\right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx

Here let's use the property

\int e^x(f(x)+f'(x))dx=e^xf(x)

so,

=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx

=\left [ -e^xcot\frac{x}{2} \right ]_\frac{\pi}{2}^\pi

=\left [ -e^\pi cot\frac{\pi}{2} \right ]-\left [ -e^{\frac{\pi}{2}} cot\frac{\pi}{4} \right ]

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

 

Posted by

Pankaj Sanodiya

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