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Mathematics Question on Integrals of Some Particular Functions

∫ex(cosec−1x+−1xx2−1) dx \int e^{x}\left(cosec^{-1}x+\frac{-1}{x\sqrt{x^{2}-1}}\right) \, dx is equal to

A

excosec−1x+Ce^x cosec^{-1} x + C

B

exsin−1x+Ce^x sin^{-1} x + C

C

exsec−1x+Ce^x sec^{-1} x + C

D

excos−1x+Ce^x cos^{-1} x + C

Answer

excosec−1x+Ce^x cosec^{-1} x + C

Explanation

Solution

∫ex(cosec−1x+−1xx2−1)dx\int e^{x} \left(cosec^{-1} x+\frac{-1}{x\sqrt{x^{2}-1}}\right) dx
=∫excosec−1 x dx−∫exxx2−1dx=\int e^{x} cosec^{-1}\,x\,dx-\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx
=[cosec−1x⋅ex−∫−1xx2−1exdx]=\left[cosec^{-1}x\cdot e^{x}-\int \frac{-1}{x\sqrt{x^{2}-1}} e^{x}dx\right]
−∫exxx2−1dx-\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx
=ex⋅cosec−1x+∫exxx2−1dx=e^{x}\cdot cosec^{-1}x+\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx
−∫exx2−1⋅x dx-\int \frac{e^{x}}{\sqrt{x^{2}-1}}\cdot x\,dx
=ex⋅cosec−1x+c=e^{x}\cdot cosec^{-1}x+c