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

& \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{aligned}$$ Hence, the integration of the expression $$\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.