Question

Integrate \[I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\phi }{{1 + \sin \phi }}} d\phi \]

Answer 1

Hint: Here first we will apply the following property of definite integrals:
\[\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} \] and some basic identities like:
\[\sin \left( {\pi - \theta } \right) = \sin \theta \]
\[1 - \left( {{{\sin }^2}\theta } \right) = {\cos ^2}\theta \]
 then solve the integral further by putting the lower and upper limit to get the exact value of the given integral.

Complete step-by-step answer:
We are given:
\[I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\phi }{{1 + \sin \phi }}} d\phi \]………………………….(1)
Applying the following property of definite integrals:
\[\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} \]
We get:-
\[I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\dfrac{{3\pi }}{4} + \dfrac{\pi }{4} - \phi } \right)}}{{1 + \sin \left( {\dfrac{{3\pi }}{4} + \dfrac{\pi }{4} - \phi } \right)}}} d\phi \]
Solving it further we get:-
\[
  I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\dfrac{{3\pi + \pi }}{4} - \phi } \right)}}{{1 + \sin \left( {\dfrac{{3\pi + \pi }}{4} - \phi } \right)}}} d\phi \\
  I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\pi - \phi } \right)}}{{1 + \sin \left( {\pi - \phi } \right)}}} d\phi \\
 \]
Now we know that:
\[\sin \left( {\pi - \theta } \right) = \sin \theta \]
Therefore applying this identity we get:-
\[I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\left( {\pi - \phi } \right)}}{{1 + \sin \phi }}} d\phi \]
Now splitting the above quantity we get:-
\[I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\pi }{{1 + \sin \phi }}} d\phi - \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\phi }{{1 + \sin \phi }}} d\phi \]
Now from equation 1 we know that :
\[I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\phi }{{1 + \sin \phi }}} d\phi \]
Therefore replacing the value in above equation we get:-
\[
  I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\pi }{{1 + \sin \phi }}} d\phi - I \\
   \Rightarrow 2I = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{\pi }{{1 + \sin \phi }}} d\phi \\
   \Rightarrow I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{1}{{1 + \sin \phi }}} d\phi \\
 \]
Now rationalizing the quantity inside the integral i.e, multiplying the denominator as well as the numerator by the conjugate of the denominator we get:-
Since the conjugate of \[1 + \sin \phi \] is \[1 - \sin \phi \]
Therefore on multiplying we get:-
\[I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{1}{{1 + \sin \phi }} \times \dfrac{{1 - \sin \phi }}{{1 - \sin \phi }}} d\phi \]
Now applying the following identity:
\[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]
 We get:-
\[
  I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{1 - \sin \phi }}{{{{\left( 1 \right)}^2} - {{\left( {\sin \phi } \right)}^2}}}} d\phi \\
  I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{1 - \sin \phi }}{{1 - \left( {{{\sin }^2}\phi } \right)}}} d\phi \\
 \]
Now we know that:
\[1 - \left( {{{\sin }^2}\theta } \right) = {\cos ^2}\theta \]
Hence applying this identity we get:-
\[I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{1 - \sin \phi }}{{{{\cos }^2}\phi }}} d\phi \]
Now splitting the quantity in integral we get:-
\[I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{1}{{{{\cos }^2}\phi }}} d\phi - \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\dfrac{{\sin \phi }}{{\cos \phi .\cos \phi }}} d\phi \]
Now we know that:-
\[
  \sec \theta = \dfrac{1}{{\cos \theta }} \\
  \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \\
 \]
Therefore substituting the values we get:-
\[I = \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {{{\sec }^2}\phi {\text{ }}} d\phi - \dfrac{\pi }{2}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} {\tan \phi \sec \phi {\text{ }}} d\phi \]
Now we will use the following known values of integrals:
\[
  \int {{{\sec }^2}\theta d\theta = \tan \theta } \\
  \int {\tan \theta \sec \theta d\theta } = \sec \theta \\
 \]
Hence we get:-
\[I = \dfrac{\pi }{2}\left[ {\tan \phi } \right]_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}} - \dfrac{\pi }{2}\left[ {\sec \phi } \right]_{\dfrac{\pi }{4}}^{\dfrac{{3\pi }}{4}}\]
Now evaluating the values by putting the upper and lower limits we get:-
\[I = \dfrac{\pi }{2}\left[ {\tan \left( {\dfrac{{3\pi }}{4}} \right) - \tan \left( {\dfrac{\pi }{4}} \right)} \right] - \dfrac{\pi }{2}\left[ {\sec \left( {\dfrac{{3\pi }}{4}} \right) - \sec \left( {\dfrac{\pi }{4}} \right)} \right]\]
Now we know that:
\[
  \tan \left( {\dfrac{{3\pi }}{4}} \right) = - 1 \\
  \tan \left( {\dfrac{\pi }{4}} \right) = 1 \\
  \sec \left( {\dfrac{{3\pi }}{4}} \right) = - \sqrt 2 \\
  \sec \left( {\dfrac{\pi }{4}} \right) = \sqrt 2 \\
 \]
Putting the respective values we get:-
\[
  I = \dfrac{\pi }{2}\left[ {\left\{ { - 1 - 1} \right\} - \left\{ { - \sqrt 2 - \sqrt 2 } \right\}} \right] \\
  I = \dfrac{\pi }{2}\left[ { - 2 + 2\sqrt 2 } \right] \\
 \]
Taking 2 as common from numerator we get:-
\[
  I = \dfrac{{2\pi }}{2}\left[ { - 1 + \sqrt 2 } \right] \\
  I = \pi \left( {\sqrt 2 - 1} \right) \\
 \]

Hence, the value of the given integral is \[\pi \left( {\sqrt 2 - 1} \right)\]


Note: Students may not use the main property of definite integrals and proceed directly but in such questions these properties should be used to simplify the question.
Also the student should substitute the values properly and use the identities used in the solutions to get the desired answer.