Question

What is the value of $\int {\left[ {\dfrac{{\left( {1 + x} \right){e^x}}}{{{{\sin }^2}\left( {x{e^x}} \right)}}} \right]} dx$?
1. $ - \cot \left( {{e^x}} \right) + c$
2. $\tan \left( {x{e^x}} \right) + c$
3. $\tan \left( {{e^x}} \right) + c$
4. $\cot \left( {x{e^x}} \right) + c$
5. $ - \cot \left( {x{e^x}} \right) + c$

Answer 1

Hint: In this question, we have to integrate the function $\dfrac{{\left( {1 + x} \right){e^x}}}{{{{\sin }^2}\left( {x{e^x}} \right)}}$ with respect to $x$. First step is to let the angle of the trigonometric term be any constant and differentiate that equation with respect to $x$. Now put the required term in the given function and using trigonometric integration formula solve further.

Formula used:
Product rule –
$\dfrac{d}{{dx}}\left( {f\left( x \right) \times g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)$
Integration formula –
$\int {\cos e{c^2}xdx = - \cot x + c} $

Complete step by step solution:
Given that,
$\int {\left[ {\dfrac{{\left( {1 + x} \right){e^x}}}{{{{\sin }^2}\left( {x{e^x}} \right)}}} \right]} dx - - - - - \left( 2 \right)$
Let, $x{e^x} = p$
Differentiate above equation with respect to $x$
$x{e^x} + {e^x} = \dfrac{{dp}}{{dx}}$
$\left( {1 + x} \right){e^x} = \dfrac{{dp}}{{dx}}$
$ \Rightarrow \left( {1 + x} \right){e^x}dx = dp - - - - - \left( 2 \right)$
From equation (1) and (2),
$\int {\left[ {\dfrac{{\left( {1 + x} \right){e^x}}}{{{{\sin }^2}\left( {x{e^x}} \right)}}} \right]} dx = \int {\dfrac{{dp}}{{{{\sin }^2}p}}} $
$ = {\int {\cos ec} ^2}pdp$
$ = - \cot p + c$
$ = - \cot \left( {x{e^x}} \right) + c$
Hence, option (5) is the correct answer i.e., $ - \cot \left( {x{e^x}} \right) + c$.

Note: The key concept involved in solving this problem is the good knowledge of integration. Students must remember that to solve any function we have to start by taking the angle or sub part of the function equal to constant and then differentiate that to make the integration function easier to solve.