Question

The value of $\int{{{e}^{\tan \theta }}\left( \sec \theta -\sin \theta \right)}\text{ }d\theta$ is equal to
(a) $-{{e}^{\tan \theta }}\sin \theta +C$
(b) ${{e}^{\tan \theta }}\sin \theta +C$
(c) ${{e}^{\tan \theta }}\sec \theta +C$
(d) ${{e}^{\tan \theta }}\cos \theta +C$

Answer 1

In this type of question we have to use the concept of integration by parts. We know that when we have to integrate a product of two functions at that time we use integration by parts. We can write the formula of integration of by parts as, $\int{u\centerdot vdx=u\int{vdx-\int{\left[ \dfrac{du}{dx}\int{vdx} \right]dx}}}$. In this case we have to select the first function that is $u$ in such a way that the derivative of the function could be easily integrated. In the given example we first separate out the integral over subtraction and then we will use the formula of integration by parts.

Complete step by step answer:
Now we have to find the value of the integral $\int{{{e}^{\tan \theta }}\left( \sec \theta -\sin \theta \right)}\text{ }d\theta$.
Let us first separate out the integral over subtraction

& \Rightarrow \int{{{e}^{\tan \theta }}\left( \sec \theta -\sin \theta \right)}\text{ }d\theta \\\ & \Rightarrow \int{{{e}^{\tan \theta }}\sec \theta d\theta -\int{{{e}^{\tan \theta }}\sin \theta d\theta }} \\\ \end{aligned}$$ Now, here we get two integrals to simplify it further. We kept the first integral as it is and we will evaluate the second integral by using the integration by parts. We know that the formula of integration by parts is given by, $$\Rightarrow \int{u\centerdot vdx=u\int{vdx-\int{\left[ \dfrac{du}{dx}\int{vdx} \right]dx}}}$$ In the second integral let us consider $$u={{e}^{\tan \theta }},v=\sin \theta $$. Thus by using above formula we can write, $$\Rightarrow \int{{{e}^{\tan \theta }}\sec \theta d\theta -\left[ {{e}^{\tan \theta }}\int{\sin \theta d\theta -\int{\left( \dfrac{d}{d\theta }{{e}^{\tan \theta }}\int{\sin \theta d\theta } \right)\text{ }}d\theta } \right]}$$ As we know that, $$\int{\sin \theta d\theta =-\cos \theta }$$ and $$\dfrac{d}{d\theta }\left( {{e}^{\tan \theta }} \right)={{e}^{\tan \theta }}{{\sec }^{2}}\theta $$ $$\Rightarrow \int{{{e}^{\tan \theta }}\sec \theta d\theta -\left[ {{e}^{\tan \theta }}\left( -\cos \theta \right)-\int{{{e}^{\tan \theta }}{{\sec }^{2}}\theta \left( -\cos \theta \right)d\theta } \right]}$$ Also we know that, $$\sec \theta =\dfrac{1}{\cos \theta }$$. Hence, by simplifying the expression further we can write, $$\begin{aligned} & \Rightarrow \int{{{e}^{\tan \theta }}\sec \theta d\theta -\left[ -{{e}^{\tan \theta }}\cos \theta +\int{{{e}^{\tan \theta }}\sec \theta d\theta } \right]} \\\ & \Rightarrow \int{{{e}^{\tan \theta }}\sec \theta d\theta +{{e}^{\tan \theta }}\cos \theta -\int{{{e}^{\tan \theta }}\sec \theta d\theta }} \\\ \end{aligned}$$ Here, we can observe that first and third terms are equal having opposite signs so they get cancel with each other $$\Rightarrow {{e}^{\tan \theta }}\cos \theta +C$$ Hence, the value of $$\int{{{e}^{\tan \theta }}\left( \sec \theta -\sin \theta \right)}\text{ }d\theta $$ is $${{e}^{\tan \theta }}\cos \theta +C$$ **So, the correct answer is “Option d”.** **Note:** In this type of question students have to remember to first separate the integral over subtraction and then use the formula of integration by parts. Also students have to note that if trigonometric functions are present and integration by parts is applicable then that trigonometric function always appears as a second function. Students have to take care in calculation of derivatives of $$\left( {{e}^{\tan \theta }} \right)$$.