Question

Evaluate the following integral
$\int{{{e}^{x}}\left( \dfrac{1+\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \right)}dx$
(a) $\dfrac{{{e}^{x}}}{\sqrt{1-{{x}^{2}}}}+c$
(b) ${{e}^{x}}{{\sin }^{-1}}x+c$
(c) ${{e}^{x}}\left( {{e}^{{{\sin }^{-1}}x}}+\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right)+c$
(d) $\left( {{e}^{{{\sin }^{-1}}x}}+\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right)+c$

Answer 1

We solve this problem by using one of the standard formulas of the integration. Whenever we see there is $'{{e}^{x}}'$ in the integration then we need to convert the function in the form
${{e}^{x}}\left( f\left( x \right)+{f}'\left( x \right) \right)$ because we have the standard formula for this type of function that is
$\int{{{e}^{x}}\left[ f\left( x \right)+{f}'\left( x \right) \right]dx={{e}^{x}}f\left( x \right)+c}$
By using the above formula we evaluate the given integral.

Complete step-by-step answer:
We are given the integral function as
$\int{{{e}^{x}}\left( \dfrac{1+\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \right)}dx$
Let us assume that the value of above integral as
$\Rightarrow I=\int{{{e}^{x}}\left( \dfrac{1+\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \right)}dx$
Here, we can see that the integral value has ${{e}^{x}}$ in it.
So, let us try to convert the integral in the form of ${{e}^{x}}\left( f\left( x \right)+{f}'\left( x \right) \right)$
Now, by dividing the terms inside the bracket of above integral we get

& \Rightarrow I=\int{{{e}^{x}}\left( \dfrac{1}{\sqrt{1-{{x}^{2}}}}+\dfrac{\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \right)}dx \\\ & \Rightarrow I=\int{{{e}^{x}}\left( \dfrac{1}{\sqrt{1-{{x}^{2}}}}+{{\sin }^{-1}}x \right)}dx.......equation(i) \\\ \end{aligned}$$ Let us assume that $$\Rightarrow f\left( x \right)={{\sin }^{-1}}x$$ Now by differentiating with respect to $$'x'$$ on both sides we get $$\Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)$$ We know that the standard formula derivative that is $$\Rightarrow \dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$$ By using this formula we get $$\Rightarrow {f}'\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$$ Now, by substituting the required values in the equation (i) we get $$\Rightarrow I=\int{{{e}^{x}}\left( f\left( x \right)+{f}'\left( x \right) \right)}dx$$ We know that the standard formula for this type of function that is $$\int{{{e}^{x}}\left[ f\left( x \right)+{f}'\left( x \right) \right]dx={{e}^{x}}f\left( x \right)+c}$$ By using the above formula we get the given integral as $$\Rightarrow I={{e}^{x}}f\left( x \right)+c$$ By substituting the value of $$f\left( x \right)$$ in above equation we get $$\Rightarrow I={{e}^{x}}{{\sin }^{-1}}x+c$$ Therefore the value of given integral is $$\therefore \int{{{e}^{x}}\left( \dfrac{1+\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \right)dx}={{e}^{x}}{{\sin }^{-1}}x+c$$ So, option (b) is the correct answer. **So, the correct answer is “Option (b)”.** **Note:** Students may make mistakes in writing the answer. Whenever we complete an integral we get some constant in the answer that is $$\int{f\left( x \right)dx}=g\left( x \right)+c$$ Here, the constant $$'c'$$ is very important. Students may miss this constant and gives the answer which will be a blunder mistake. It may not have significance in this problem but it has more significance in other problems. So we should not miss this constant after the integral evaluation. For this type of question, students must remember the formula, $$\int{{{e}^{x}}\left[ f\left( x \right)+{f}'\left( x \right) \right]dx={{e}^{x}}f\left( x \right)+c}$$ which helps a lot in solving the question quickly, especially for competitive exams. So, memorising such formulas will help a lot.