Question

How do you evaluate\[\int {\dfrac{{\arcsin x}}{{\sqrt {1 + x} }}dx} \]?

Answer 1

Hint: Since we know that integration is the reverse process of differentiation. The above given problem can be solved by use of some basic integration and differentiation. Here the method implied will be integration by parts. Lastly we can use substitution technique for solving questions where multiple terms are present as product terms inside integration.

Formula used : In order to solve the given problem we will use integration by parts method. The formula of integration by parts method is
\[\int {udv = uv - \int {vdu} } \]
\[
   \Rightarrow u = \arcsin (x) \\
   \Rightarrow dv = \dfrac{{dx}}{{\sqrt {1 + x} }} \\
   \Rightarrow du = \dfrac{1}{{\sqrt {1 - {x^2}} }}dx \\
   \Rightarrow dv = \dfrac{{dx}}{{\sqrt {1 + x} }} \\
   \Rightarrow \int {dv} = \int {\dfrac{{dx}}{{\sqrt {1 + x} }}} \\
   \Rightarrow v = 2\sqrt {1 + x} \\
 \]
By applying this in the given equation we can evaluate it to get the desired results.

Complete step by step answer:
Here as we know that we are asked to evaluate \[\int {\dfrac{{\arcsin x}}{{\sqrt {1 + x} }}dx} \]
Since we know that \[\int {udv = uv - \int {vdu} } \]
\[
   \Rightarrow u = \arcsin (x) \\
   \Rightarrow dv = \dfrac{{dx}}{{\sqrt {1 + x} }} \\
 \]
Now secondly after differentiating \[u = \arcsin x\]we get
\[du = \dfrac{1}{{\sqrt {1 - {x^2}} }}dx\]
Then integrate \[dv = \dfrac{{dx}}{{\sqrt {1 + x} }}\]
\[
   \Rightarrow \int {dv} = \int {\dfrac{{dx}}{{\sqrt {1 + x} }}} \\
   \Rightarrow v = 2\sqrt {1 + x} \\
 \]
Therefore we get
\[\arcsin (x)(2\sqrt {1 + x} ) - \int 2 \sqrt {1 + x} \left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)dx\]
Now after rewriting and factor \[1 - {x^2}\]
\[
   \Rightarrow 2\arcsin (x)\sqrt {1 + x} - 2\int {\dfrac{{\sqrt {1 + x} }}{{\sqrt {(1 + x)(1 - x)} }}dx} \\
   \Rightarrow 2\arcsin (x)\sqrt {1 + x} - 2\int {\dfrac{{\sqrt {1 + x} }}{{\sqrt {1 + x} \sqrt {1 - x} }}dx} \\
   \Rightarrow 2\arcsin (x)\sqrt {1 + x} - 2\int {\dfrac{1}{{\sqrt {1 - x} }}dx} \\
 \]
Finally we will integrate and we get
\[2\arcsin (x)\sqrt {1 + x} + 4\sqrt {1 - x} + C\]
Here we will find an additional term \[C\] known as integration constant which we get while performing indefinite integration.

Note: For solving such types of questions we should have a good grasp of integration formula. One should recall the definition of tangent in terms of sine and cosine. By applying the right technique most of the problems can be solved easily so always remember the shorthand methods for multiple applications for integration by parts problems. Integrate carefully and by applying the required formula to get the desired result.