Question

How to integrate \[\arcsin x\] with respect to $x$?

Answer 1

Hint:
In the given question, we have been given a trigonometric function. We have to integrate it. To do that, we first simplify the value of the expression by using substitution. Then we use integration by substitution and integration by parts to evaluate the answer to the given question.

Complete step by step answer:
First, we are going to make some substitution:
Let \[t = \arcsin x \Rightarrow x = \sin \left( t \right)\] and \[\dfrac{{dx}}{{dt}} = \cos \left( t \right)\]
Substituting the values into the main expression,
\[\int {\arcsin \left( x \right)dx = \int {t\cos \left( t \right)dt} } \]
Let \[u = t\] and \[dv = \cos \left( t \right)dt\]
Then \[du = dt\] and \[v = \sin \left( t \right)\]
Using the formula of integration by parts,
\[\int {t\cos \left( t \right)dt = t\sin \left( t \right) - \int {\sin \left( t \right)dt} } \]
Solving,
\[ = t\sin \left( t \right) + \cos \left( t \right) + C\]
Substituting back the original values,
\[ = \arcsin \left( x \right).\sin \left( {\arcsin \left( x \right)} \right) + \cos \left( {\arcsin \left( x \right)} \right) + C\]
Now, we know that
\[\sin \left( {\arcsin \left( x \right)} \right) = x\] and \[\cos \left( {\arcsin \left( x \right)} \right) = \sqrt {1 - {x^2}} \]
Thus,
\[\int {\arcsin \left( x \right)} dx = x\arcsin \left( x \right) + \sqrt {1 - {x^2}} + C\]

Additional Information:
Integration is the opposite of differentiation. In differentiation, we “break” things for examining how they behave separately. While, in integration, we combine the expressions so as to see their collective behavior. If we have a definite integral, then we calculate its value by putting in the upper limit into the result, then putting in the lower limit into the result, and then subtracting the two. A definite integral is the one which looks like,
\[\int\limits_b^a \text{some }\text{expression} \]

Note:
In the given question, we had to find the integral of arcsin function. We did that by first substituting the value of the function with a variable. Then we proceeded by using integration by substitution and then using integration by parts. It is very important that we know the exact formula and, where, when, and how to apply it.