Question

The value of $I = \smallint _{\pi /2}^{5\pi /2}\dfrac{{{e^{{{\tan }^{ - 1}}(\sin x)}}}}{{{e^{{{\tan }^{ - 1}}(\sin x)}} + {e^{{{\tan }^{ - 1}}(\cos x)}}}}dx$ is
A. 1
B. $\pi$
C. $e$
D. $\dfrac{\pi }{2}$

Answer 1

Hint: Split the integral into two different integrals using limits. And then use property of definite integrals to evaluate the integral.

Complete step-by-step answer:
According to the question, the given integral is:
$\Rightarrow I = \smallint _{\pi /2}^{5\pi /2}\dfrac{{{e^{{{\tan }^{ - 1}}(\sin x)}}}}{{{e^{{{\tan }^{ - 1}}(\sin x)}} + {e^{{{\tan }^{ - 1}}(\cos x)}}}}dx$
The limit of the integration is from the $\dfrac{\pi }{2}$ to $\dfrac{{5\pi }}{2}$. According to the property of definite integral, we can split the limit into two parts i.e. from $\dfrac{\pi }{2}$ to $\pi$ and from $\pi$ to $\dfrac{{5\pi }}{2}$. Doing so, we’ll get:$\Rightarrow I = \smallint _{\pi /2}^\pi \dfrac{{{e^{{{\tan }^{ - 1}}(\sin x)}}}}{{{e^{{{\tan }^{ - 1}}(\sin x)}} + {e^{{{\tan }^{ - 1}}(\cos x)}}}}dx + \smallint _\pi ^{5\pi /2}\dfrac{{{e^{{{\tan }^{ - 1}}(\sin x)}}}}{{{e^{{{\tan }^{ - 1}}(\sin x)}} + {e^{{{\tan }^{ - 1}}(\cos x)}}}}dx .....(i)$
Now, another property of definite integral is:
$\Rightarrow \smallint _a^bf(x)dx = \smallint _a^bf(a + b - x)dx$.
Using this property for the above integral, we’ll get:
$\Rightarrow$$$I = \smallint _{\pi /2}^\pi \dfrac{{{e^{{{\tan }^{ - 1}}\left( {\sin \left( {3\pi /2 - x} \right)} \right)}}}}{{{e^{{{\tan }^{ - 1}}\left( {\sin \left( {3\pi /2 - x} \right)} \right)}} + {e^{{{\tan }^{ - 1}}\left( {\cos \left( {3\pi /2 - x} \right)} \right)}}}}dx + \smallint _\pi ^{5\pi /2}\dfrac{{{e^{{{\tan }^{ - 1}}\left( {\sin \left( {7\pi /2 - x} \right)} \right)}}}}{{{e^{{{\tan }^{ - 1}}\left( {\sin \left( {7\pi /2 - x} \right)} \right)}} + {e^{{{\tan }^{ - 1}}\left( {\cos \left( {7\pi /2 - x} \right)} \right)}}}}dx$We know that$\sin \left( {3\pi /2 - x} \right) = \cos x,\cos \left( {3\pi /2 - x} \right) = \sin x,\sin \left( {7\pi /2 - x} \right) = \cos x$and$\cos \left( {7\pi /2 - x} \right) = \sin x$. Using these values, we’ll get:$I = \smallint _{\pi /2}^\pi \dfrac{{{e^{{{\tan }^{ - 1}}\left( {\cos x} \right)}}}}{{{e^{{{\tan }^{ - 1}}\left( {\cos x} \right)}} + {e^{{{\tan }^{ - 1}}\left( {\sin x} \right)}}}}dx + \smallint _\pi ^{5\pi /2}\dfrac{{{e^{{{\tan }^{ - 1}}\left( {\cos x} \right)}}}}{{{e^{{{\tan }^{ - 1}}\left( {\cos x} \right)}} + {e^{{{\tan }^{ - 1}}\left( {\sin x} \right)}}}}dx .....(ii)$Adding equation$(i)$and$(ii)$$, we’ll get:

$$\Rightarrow 2I = \smallint _{\pi /2}^\pi \dfrac{{{e^{{{\tan }^{ - 1}}(\sin x)}} + {e^{{{\tan }^{ - 1}}(\cos x)}}}}{{{e^{{{\tan }^{ - 1}}(\sin x)}} + {e^{{{\tan }^{ - 1}}(\cos x)}}}}dx + \smallint _\pi ^{5\pi /2}\dfrac{{{e^{{{\tan }^{ - 1}}(\sin x)}} + {e^{{{\tan }^{ - 1}}(\cos x)}}}}{{{e^{{{\tan }^{ - 1}}(\sin x)}} + {e^{{{\tan }^{ - 1}}(\cos x)}}}}dx \\\ \Rightarrow 2I = \smallint _{\pi /2}^\pi dx + \smallint _\pi ^{5\pi /2}dx \\\$$ $$\Rightarrow 2I = \left[ x \right]_{\pi /2}^\pi + \left[ x \right]_\pi ^{5\pi /2} \\\ \Rightarrow 2I = \pi - \dfrac{\pi }{2} + \dfrac{{5\pi }}{2} - \pi \\\ \Rightarrow 2I = 2\pi \\\ \Rightarrow I = \pi \\\$$

Thus the value of integral is $\pi$.

Note: We can split a definite integral up into two integrals with the same integrand but different limits. Integration can be interpreted as the area under the curve within a particular limit. By splitting the integral into two parts means we are splitting the areas under the curve into two different continuous limits. The value of the integral is the sum of two areas.