Question

Evaluate \[\int {\sqrt {1 + {y^2}} \cdot 2ydy} \]
A) \[I = \dfrac{2}{3}{\left( {1 + {y^2}} \right)^{3/2}} + C\]
B) \[I = \dfrac{2}{5}{\left( {1 - {y^2}} \right)^{3/2}} + C\]
C) \[I = \dfrac{2}{3}{\left( {1 - {y^2}} \right)^{3/2}} + C\]
D) None of these

Answer 1

Hint:
Here we will firstly substitute the equation inside the square root to some variable and find its differentiation. Then we will substitute the variable and its derivative in the given equation. Then we will use the concept of the integration to integrate the equation. Then we will put the actual value of the function in place of the substituted variable in the integrated equation to get final integration value.

Formula used:
We will use the formula of integration, \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} \], to solve the question.

Complete step by step solution:
Let \[I = \int {\sqrt {1 + {y^2}} \cdot 2ydy} \].
Let us now assume \[u = 1 + {y^2}\].
Now we will differentiate \[u\].
Differentiating both sides of the above equation, we get
  \[ \Rightarrow du = 2y \cdot dy\]
Now substituting \[u = 1 + {y^2}\] and \[du = 2y \cdot dy\] in \[I = \int {\sqrt {1 + {y^2}} \cdot 2ydy} \], we get
 \[ \Rightarrow I = \int {{u^{\dfrac{1}{2}}}} du\]
Integrating the terms using the formula \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} \], we get
 \[ \Rightarrow I = \dfrac{{{u^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}} + C\]
Adding the terms in the denominator and the exponent, we get
 \[ \Rightarrow I = \dfrac{2}{3}{u^{\dfrac{3}{2}}} + C\]
Now we will put the value of \[u\] in the above equation to get the value of the integration in terms of the \[y\]. Therefore, we get
 \[ \Rightarrow I = \dfrac{2}{3}{\left( {1 + {y^2}} \right)^{\dfrac{3}{2}}} + C\]
Hence the value of the \[\int {\sqrt {1 + {y^2}} \cdot 2ydy} \] is equal to \[I = \dfrac{2}{3}{\left( {1 + {y^2}} \right)^{3/2}} + C\].

So, option A is the correct option.

Note:
Here we should note that the value of the substitution must be done accordingly. In the integration with the limits we also have to convert the limits according to the substitution but in our case it's simple integration without limits of the integration. Differentiation is the opposite of the integration i.e. differentiation of the integration is equal to the value of the function or vice versa. We have to remember to put the constant term \[C\] after the integration of an equation.