Question:
$ \int e^{x}\left(cosec^{-1}x+\frac{-1}{x\sqrt{x^{2}-1}}\right) \, dx$ is equal to
$ \int e^{x}\left(cosec^{-1}x+\frac{-1}{x\sqrt{x^{2}-1}}\right) \, dx$ is equal to
Updated On: Jun 14, 2022
- $ e^x cosec^{-1} x + C $
- $ e^x sin^{-1} x + C $
- $ e^x sec^{-1} x + C $
- $ e^x cos^{-1} x + C $
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The Correct Option is A
Solution and Explanation
$\int e^{x} \left(cosec^{-1} x+\frac{-1}{x\sqrt{x^{2}-1}}\right) dx$
$=\int e^{x} cosec^{-1}\,x\,dx-\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx$
$=\left[cosec^{-1}x\cdot e^{x}-\int \frac{-1}{x\sqrt{x^{2}-1}} e^{x}dx\right]$
$-\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx$
$=e^{x}\cdot cosec^{-1}x+\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx $
$-\int \frac{e^{x}}{\sqrt{x^{2}-1}}\cdot x\,dx$
$=e^{x}\cdot cosec^{-1}x+c$
$=\int e^{x} cosec^{-1}\,x\,dx-\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx$
$=\left[cosec^{-1}x\cdot e^{x}-\int \frac{-1}{x\sqrt{x^{2}-1}} e^{x}dx\right]$
$-\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx$
$=e^{x}\cdot cosec^{-1}x+\int \frac{e^{x}}{x\sqrt{x^{2}-1}}dx $
$-\int \frac{e^{x}}{\sqrt{x^{2}-1}}\cdot x\,dx$
$=e^{x}\cdot cosec^{-1}x+c$
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
