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
⇒ \(∫\frac {1}{\sqrt {(4)^2-(x+3)^2}} dx\) = \(∫\frac {1}{\sqrt {(4)^2-t^2}} dt\)
= \(sin^{-1}(\frac t4)+C\)
=\(sin^{-1}(\frac {x+3}4)+C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
Examine Bernier's opinion on the question of land ownership in Mughal India and how were the western economists influenced by Bernier's description?
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.
These are tabulated below along with the meaning of each part.