Antiderivative Of Arcsin
Antiderivative Of Arcsin
We know that the inverse operations of differentiation is said to be anti derivatives. It
is also related to definite integrals during the first fundamental theorem of calculus.
And the representation of anti derivatives is f (p) on the function f (p) which is known
as integrand, and the function is anti differentiated or integrated.
And the anti derivative of a function f (p) is another function F (p) which is having the
derivative F' (p) = f (p), and the generalized form of anti derivative of f (p) with respect
to the p is f (p) = F (p) + C;
Where, the value of `C' is known as the constant of integration. This formula is
applicable because the value of constant term is zero.
For example: p2 is the anti derivative of 2p because 2p is the derivative of p2.
Now we will see how to find the antiderivative of arcsin.For finding the anti derivative
of arcsin we have to follow some of the steps which are given below:
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Step1: First it is necessary to know about what the arcsin is?
Step2: we know that the arcsin is the inverse function of sine. It is represented as sin-
1. Suppose we have p = sin q, then value of ' q' will be equals to sin -1 p or it can be
written as q = sin -1 p or q = arcsine p.
Step3: Then we will find the anti derivative of arcsine p, in this it is necessary to find
the integral of arcsine p.
As we know that the definition of anti derivative, anti derivative is operation opposite
to differentiation operation which is Integration.
Step4: Then we will find anti derivative of arcsine p. The process of calculation of anti
derivative of arcsine p is given below-

arcsin p dp;
Let the value of a = arcsin (p)
Therefore da = 1 / (1 - p2) dp;
Let db = dp, which gives b = p after integration.
Now apply process of Integration by parts:
a. db = a.b - b . da
arcsin (p) . p - [(p). 1 / (1 - p2)].dp .......................... (1)
ReadMoreAboutAntiderivativeInverseTangent


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Now we will substitute with t- Let t = 1 - p2
Therefore dt = - 2p. dp
dt / (- 2p) = dp
Substitute 1 - p2 = t and dt / (-2p) = dp, so we get-
Arcsin (p) .p - [(p). 1 / (t). dt / (-2p)];
On further solving the above expression we get:
Arcsin (p) .p - (- 1/2) [1 / (t). dt]
Arcsin (p) .p + (1/2) [(t)-1/2 + 1 / (- 1/2 + 1)] + c
Arcsin (p) .p + (1/2) [(t) 1/2 / (2 / 1)] + c
Arcsin (p) .p + (t) 1/2 + c
Again substitute value of't ' we get-
arcsin (p) * p + (1-p) + C
When we solve the expression we get the expression:
Hence antiderivative of arcsin p is arcsin (p) * p + (1-p) + C.
This is how to solve the anti derivatives of arcsin. This is important concept of
mathematics.


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