Question

Integrate the given expression, \[\int{{{\tan }^{-1}}\left( \sqrt{\dfrac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)}} \right)}dx\] .

Answer 1

Hint:We know the formula, \[\sqrt{1-\sin 2x}=\cos x-\sin x\] . We also know the formula, \[\sqrt{1+\sin 2x}=\sin x+\cos x\] . We don’t have 2x terms in the expression. So, we have to replace x by \[\dfrac{x}{2}\] in both formulas. We get \[\sqrt{1-\sin x}=\cos \dfrac{x}{2}-\sin \dfrac{x}{2}\] and \[\sqrt{1+\sin x}=\sin \dfrac{x}{2}+\cos \dfrac{x}{2}\] . Now, substituting this in the given expression we get, \[\int{{{\tan }^{-1}}\left( \sqrt{\dfrac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)}} \right)}dx=\int{{{\tan }^{-1}}\left( \dfrac{\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}+\sin \dfrac{x}{2}} \right)}\] . Now divide by \[\cos \dfrac{x}{2}\] in numerator and denominator of the expression and then use the formula, \[\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}\] , where A = \[\dfrac{\pi }{4}\] and B = \[\dfrac{x}{2}\] . Use the property, \[{{\tan }^{-1}}(tanx)=x\] and solve further.

Complete step-by-step answer:
According to the question, we have to integrate the expression \[\int{{{\tan }^{-1}}\left( \sqrt{\dfrac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)}} \right)}dx\] . We have to convert this expression into simpler form.
We know the formula,
\[\sqrt{1-\sin 2x}=\cos x-\sin x\] …………………….(1)
\[\sqrt{1+\sin 2x}=\sin x+\cos x\] ……………………..(2)
Replacing x by \[\dfrac{x}{2}\] in equation (1) and equation (2), we get
\[\sqrt{1-\sin 2.\dfrac{x}{2}}=\cos \dfrac{x}{2}-\sin \dfrac{x}{2}\]
 \[\Rightarrow \sqrt{1-\sin x}=\cos \dfrac{x}{2}-\sin \dfrac{x}{2}\] ……………………….(3)
\[\sqrt{1+\sin 2.\dfrac{x}{2}}=\sin \dfrac{x}{2}+\cos \dfrac{x}{2}\]

\[\Rightarrow \sqrt{1+\sin x}=\sin \dfrac{x}{2}+\cos \dfrac{x}{2}\] ………………….(4)
Using equation (3) and equation (4), we can transform the given expression.
\[\int{{{\tan }^{-1}}\left( \sqrt{\dfrac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)}} \right)}dx\]
\[\Rightarrow \int{{{\tan }^{-1}}\left( \dfrac{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)} \right)}dx\] ………………….(5)
Dividing by \[\cos \dfrac{x}{2}\] in numerator and denominator of the equation (5).
\[\Rightarrow \int{{{\tan }^{-1}}\left( \dfrac{\left( \dfrac{\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}}-\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}{\left( \dfrac{\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}}+\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)} \right)}dx\] ……………………..(6)
We know that, \[\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}=\tan \dfrac{x}{2}\] …………………..(7)
From equation (6) and equation (7), we have
\[\Rightarrow \int{{{\tan }^{-1}}\left( \dfrac{\left( 1-\tan \dfrac{x}{2} \right)}{\left( 1+\tan \dfrac{x}{2} \right)} \right)}dx\] ………………………….(8)
We also know the formula, \[\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}\] ………….(9)
Now, replacing A = \[\dfrac{\pi }{4}\] and B = \[\dfrac{x}{2}\] in the equation (9), we have
\[\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}\]
\[\Rightarrow \tan (\dfrac{\pi }{4}-\dfrac{x}{2})=\dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}}\] …………………………(10)
We know that, \[\tan \dfrac{\pi }{4}=1\] ………………(11)
From equation (10) and equation (11), we have
\[\Rightarrow \tan (\dfrac{\pi }{4}-\dfrac{x}{2})=\dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}}\]
\[\Rightarrow \tan (\dfrac{\pi }{4}-\dfrac{x}{2})=\dfrac{1-\tan \dfrac{x}{2}}{1+1.\tan \dfrac{x}{2}}\] …………………………(12)
From equation (8) and equation (12), we get
\[\Rightarrow \int{{{\tan }^{-1}}\left( \dfrac{\left( 1-\tan \dfrac{x}{2} \right)}{\left( 1+\tan \dfrac{x}{2} \right)} \right)}dx\]
\[\Rightarrow \int{{{\tan }^{-1}}(\tan (\dfrac{\pi }{4}-\dfrac{x}{2}))dx}\] ………………………..(13)
We know the property, \[{{\tan }^{-1}}(tanx)=x\] . Using this property in equation (13), we get
\[\begin{align}
  & \int{\left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)dx} \\
 & =\dfrac{\pi }{4}x-\dfrac{1}{2}.\dfrac{{{x}^{2}}}{2} \\
 & =\dfrac{\pi x}{4}-\dfrac{{{x}^{2}}}{4}+C \\
\end{align}\]
Hence, the integration of the expression \[\int{{{\tan }^{-1}}\left( \sqrt{\dfrac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)}} \right)}dx\] is \[\dfrac{\pi x}{4}-\dfrac{{{x}^{2}}}{4}+C\] .

Note: In equation (5), one may think to divide by \[sin\dfrac{x}{2}\] , then we will have \[\cot \dfrac{x}{2}\] in numerator and denominator of the expression. If we don’t convert \[\cot \dfrac{x}{2}\] into \[tan\dfrac{x}{2}\] . Then, we will not be able to use the formula, \[\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}\] and the property \[{{\tan }^{-1}}(tanx)=x\] . If this formula and property is not applied then our expression will become difficult to solve.