Question

What is the value of the integral \[\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\varphi }{{1 + \sin \varphi }}} d\varphi \]?
A. \[\pi \tan \dfrac{\pi }{8}\]
B. \[\log \tan \dfrac{\pi }{8}\]
C. \[\tan \dfrac{\pi }{8}\]
D. None of these

Answer 1

Hint: Here, a definite integral is given. First, simplify the function by using the definite integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]. Then, add this integral with the original integral and solve the integral. After that, multiply the numerator and denominator by \[1 - \sin \varphi \] and simplify it by using the trigonometric identities. In the end, solve the integrals by using the standard integral formulas and apply the upper and lower limits to get the required answer.

Formula Used: Definite integration rule: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]
\[{\sin ^2}x + {\cos ^2}x = 1\]
\[\int\limits_a^b {{{\sec }^2}xdx = \left[ {\tan x} \right]} _a^b\]
\[\int\limits_a^b {\sec x\tan xdx = \left[ {\sec x} \right]} _a^b\]

Complete step by step solution: The given definite integral is \[\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\varphi }{{1 + \sin \varphi }}} d\varphi \].

Let consider,
\[I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\varphi }{{1 + \sin \varphi }}} d\varphi \] \[.....\left( 1 \right)\]
Apply the definite integral rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] on the right-hand side.
\[I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\dfrac{\pi }{4} + \dfrac{{3\pi }}{4} - \varphi } \right)}}{{1 + \sin \left( {\dfrac{\pi }{4} + \dfrac{{3\pi }}{4} - \varphi } \right)}}} d\varphi \]
\[ \Rightarrow I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\dfrac{{4\pi }}{4} - \varphi } \right)}}{{1 + \sin \left( {\dfrac{{4\pi }}{4} - \varphi } \right)}}} d\varphi \]
\[ \Rightarrow I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\pi - \varphi } \right)}}{{1 + \sin \left( {\pi - \varphi } \right)}}} d\varphi \]
\[ \Rightarrow I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\pi - \varphi } \right)}}{{1 + \sin \varphi }}} d\varphi \] \[.....\left( 2 \right)\]
Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[2I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\varphi }{{1 + \sin \varphi }}} d\varphi + \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\pi - \varphi } \right)}}{{1 + \sin \varphi }}} d\varphi \]
\[ \Rightarrow 2I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\varphi + \pi - \varphi } \right)}}{{1 + \sin \varphi }}} d\varphi \]
\[ \Rightarrow 2I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\pi }{{1 + \sin \varphi }}} d\varphi \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{1}{{1 + \sin \varphi }}} d\varphi \]
Now multiply the numerator and denominator by \[1 - \sin \varphi \].
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{1 - \sin \varphi }}{{\left( {1 + \sin \varphi } \right)\left( {1 - \sin \varphi } \right)}}} d\varphi \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{1 - \sin \varphi }}{{1 - {{\sin }^2}\varphi }}} d\varphi \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{1 - \sin \varphi }}{{{{\cos }^2}\varphi }}} d\varphi \]
Simplify the function by using the trigonometric ratios.
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\left[ {\dfrac{1}{{{{\cos }^2}\varphi }} - \dfrac{{\sin \varphi }}{{{{\cos }^2}\varphi }}} \right]} d\varphi \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\left[ {\dfrac{1}{{{{\cos }^2}\varphi }} - \dfrac{{\sin \varphi }}{{\cos \varphi }}\dfrac{1}{{\cos \varphi }}} \right]} d\varphi \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\left[ {{{\sec }^2}\varphi - \tan \varphi \sec \varphi } \right]} d\varphi \]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {{{\sec }^2}\varphi } d\varphi - \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\tan \varphi \sec \varphi } d\varphi } \right]\]
Apply the standard integration formulas of the trigonometric functions.
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {\left[ {\tan \varphi } \right]_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} - \left[ {\sec \varphi } \right]_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}}} \right]\]
Apply the upper and lower limits.
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {\left( {\tan \dfrac{{3\pi }}{4} - \tan \dfrac{\pi }{4}} \right) - \left( {\sec \dfrac{{3\pi }}{4} - \sec \dfrac{\pi }{4}} \right)} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {\left( { - 1 - 1} \right) - \left( { - \sqrt 2 - \sqrt 2 } \right)} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {\left( { - 2} \right) - \left( { - 2\sqrt 2 } \right)} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ { - 2 + 2\sqrt 2 } \right]\]
\[ \Rightarrow I = \pi \left( {\sqrt 2 - 1} \right)\]
\[ \Rightarrow I = \pi \tan \dfrac{\pi }{8}\]
Therefore,
\[\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\varphi }{{1 + \sin \varphi }}} d\varphi = \pi \tan \dfrac{\pi }{8}\]


Option ‘A’ is correct

Note: Students often get confused about the formula of the definite integral of the function. They used \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) + F\left( a \right)\] , which is incorrect. The correct formula is \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].