Integrate the function: \(\frac {1}{\sqrt {7-6x-x^2}}\)
Solution and Explanation
7-6x-x2 can be written as 7-(x2+6x+9-9).
Therefore,
7-(x2+6x+9-9)
= 16-(x2+6x+9)
= 16-(x+3)2
= (4)2-(x+3)2
∴ \(∫\)\(\frac {1}{\sqrt {7-6x-x^2}}\ dx\) = \(∫\frac {1}{\sqrt {(4)^2-(x+3)^2}} dx\)
Let x+3 = t
⇒ dx = dt
⇒ \(∫\frac {1}{\sqrt {(4)^2-(x+3)^2}} dx\) = \(∫\frac {1}{\sqrt {(4)^2-t^2}} dt\)
= \(sin^{-1}(\frac t4)+C\)
=\(sin^{-1}(\frac {x+3}4)+C\)
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Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
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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.
