Question:
$\int sin^{-1} x dx$ is equal to
$\int sin^{-1} x dx$ is equal to
Updated On: Jul 6, 2022
- $cos^{- 1 }x + C$
- $ xsin^{-1} x+\sqrt{1-x^{2} }+C$
- $ \frac{1}{\sqrt{1-x^{2}}}+C$
- $ x sin^{-1}x-\sqrt{1-x^{2}}+C$
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The Correct Option is B
Solution and Explanation
Let
$I=\int sin^{-1}x \cdot 1 dx = \left(sin^{-1}x\right)x -\int\frac{1}{\sqrt{1-x^{2}}}\cdot x dx +C' $
$ Put 1-x^{2} =t^{2 }\Rightarrow -2x dx = 2t dt $
$x sin^{-1}x - \int\frac{\left(-t dt \right)}{t}+ C' = x sin^{-1} x + \sqrt{1-x^{2}}+C$
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.