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Question: Integrate the function, \(\int{{{e}^{-x}}\csc x\left( 1+\cot x \right)dx}\) (a) \({{e}^{-x}}\csc x...

Integrate the function, ∫e−xcsc⁡x(1+cot⁡x)dx\int{{{e}^{-x}}\csc x\left( 1+\cot x \right)dx}
(a) e−xcsc⁡x+C{{e}^{-x}}\csc x+C
(b) −e−xcsc⁡x+C-{{e}^{-x}}\csc x+C
(c) −e−x(csc⁡x+cot⁡x)+C-{{e}^{-x}}\left( \csc x+\cot x \right)+C
(d) −e−x(csc⁡x+tanx)+C-{{e}^{-x}}\left( \csc x+tanx \right)+C
(e) −e−xsec⁡x+C-{{e}^{-x}}\sec x+C

Explanation

Solution

Hint: First convert e−xcsc⁡x(1+cot⁡x){{e}^{-x}}\csc x\left( 1+\cot x \right) to −e−x(−(csc⁡xcot⁡x)−csc⁡x)-{{e}^{-x}}\left(- \left( \csc x\cot x \right)-\csc x \right) then apply the formula,
∫e−x(f′(x)−f(x))=e−xf(x)+C\int{{{e}^{-x}}\left( f'\left( x \right)-f\left( x \right) \right)={{e}^{-x}}f\left( x \right)+C}, here f(x) is a function and f′(x){{f}^{'}}\left( x \right) is obtained on differentiating.

Complete Step-by-Step solution:
In these types of question we have to use identity,
∫e−x(f′(x)−f(x))dx=e−xf(x)\int{{{e}^{-x}}\left( f'\left( x \right)-f\left( x \right) \right)dx={{e}^{-x}}f\left( x \right)}
Here in this case if f(x) be a function of x and f′(x)f'\left( x \right) will be a function we got after differentiation of f(x) with respect to x and it is of form e−x(f′(x)−f(x)){{e}^{-x}}\left( {{f}^{'}}\left( x \right)-f\left( x \right) \right) then after integration it will become e−xf(x){{e}^{-x}}f\left( x \right) .
In the given question, we have e−xcsc⁡x(1+cot⁡x){{e}^{-x}}\csc x\left( 1+\cot x \right) then we can write it as,
−e−x((csc⁡xcot⁡x)−csc⁡x)........(i)-{{e}^{-x}}\left( \left( \csc x\cot x \right)-\csc x \right)........(i)
Let us consider csc⁡x\csc x be f(x). Then on differentiating f(x) we get,
f′(x)=−csc⁡xcot⁡xf'\left( x \right)=-\csc x\cot x
So, we can write the expression (i) as,
−e−x(f′(x)−f(x))-{{e}^{-x}}\left( {{f}^{'}}\left( x \right)-f\left( x \right) \right)
Here f(x)=csc⁡xf\left( x \right)=\csc x.
So, applying the formula, ∫e−x(f′(x)−f(x))dx=e−xf(x)\int{{{e}^{-x}}\left( f'\left( x \right)-f\left( x \right) \right)dx={{e}^{-x}}f\left( x \right)}, the integration of −e−x((csc⁡xcot⁡x)−csc⁡x)-{{e}^{-x}}\left( \left( \csc x\cot x \right)-\csc x \right) will be represented as,
∫−e−x((−csc⁡xcot⁡x)−csc⁡x)=−e−x(csc⁡x)+C\int{-{{e}^{-x}}\left( \left( -\csc x\cot x \right)-\csc x \right)=}-{{e}^{-x}}\left( \csc x \right)+C
Therefore the value of the integration of function e−xcsc⁡x(1+cot⁡x){{e}^{-x}}\csc x\left( 1+\cot x \right) is −e−x(cscx)+C-{{e}^{-x}}\left( cscx \right)+C.
Hence, the correct answer is option (b).

Note: Student should be careful while transforming the expression e−xcsc⁡x(1+cot⁡x){{e}^{-x}}\csc x\left( 1+\cot x \right) to −e−x((−csc⁡xcot⁡x)−csc⁡x)-{{e}^{-x}}\left( \left( -\csc x\cot x \right)-\csc x \right). This is the area where students generally can’t get the idea what to do, they should visualize using practise.