Integrate $\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx$

Question

Integrate $\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx$

My Attempt

Put $x=\cos2a\implies dx=-2\sin2a.da$ $$ \int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx=\int\tan^{-1}\sqrt{\frac{1-\cos2a}{1+\cos2a}}.-2\sin2a.da\\ =-2\int\tan^{-1}\sqrt{\frac{2\sin^2a}{2\cos^2a}}.\sin2a.da=-2\int\tan^{-1}(\tan a)\sin2a.da $$ We have $y=\tan^{-1}(\tan a)\implies\tan y=\tan a\implies y=n\pi+a$ $$ \begin{align} &\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}dx=-2\int(n\pi+a)\sin2a.da=-2n\pi a-2\int a.\sin2a.da\\ &=-2n\pi a-2\bigg[a\frac{-\cos2a}{2}-\int\frac{-\cos2a}{2}da\bigg]\\ &=-2n\pi a+a.\cos2a+\frac{\sin2a}{2}+C\\ &=-2n\pi.\frac{1}{2}\cos^{-1}x+\frac{1}{2}\cos^{-1}x.x+\frac{\sqrt{1-x^2}}{2}+C\\ &\color{red}{=\frac{1}{2}\bigg[-2n\pi\cos^{-1}x+x\cos^{-1}x-\sqrt{1-x^2}\bigg]} \end{align} $$ My reference has the solution $\frac{1}{2}\bigg[x\cos^{-1}x-\sqrt{1-x^2}\bigg]$. But, why am I getting the solution as above ?

Answer 1

You need to notice that you just substituted $$a=\frac {\arccos x}{2}$$ whose range is itself $$\left[0,\frac {\pi}{2 }\right]$$.

And as you might know that for $\alpha \in \left(\frac {-\pi}{2},\frac {\pi}{2}\right)$ , $$\arctan (\tan \alpha) =\alpha$$ Hence $$\arctan (\tan a) =a$$ Therefore you won't get an $n\pi$ term

Answer 2

Utilize the identity $\tan^{-1}\sqrt{\frac{1-x}{1+x}}=\frac12\cos^{-1}x$, followed by integration by parts $$\int \tan^{-1}\sqrt{\frac{1-x}{1+x}}dx=\frac12 \int \cos^{-1}x\ dx\overset{ibp}= \frac{1}{2}\left(x\cos^{-1}x-\sqrt{1-x^2}\right) $$

Answer 3

$$\begin{align*} & \int \tan^{-1}\sqrt{\frac{1-x}{1+x}} \, dx \\ &= -4 \int \frac{y\tan^{-1}y}{(1+y^2)^2} \, dy \tag1 \\ &= \frac{2 \tan^{-1}y}{1+y^2} - 2 \int \frac{dy}{(1+y^2)^2} \tag2 \\ &= \frac{2\tan^{-1}y}{1+y^2} - \tan^{-1}y - \frac y{1+y^2} + C \tag2 \\ &= \frac{2\tan^{-1}\sqrt{\frac{1-x}{1+x}}}{1+\frac{1-x}{1+x}} - \tan^{-1}\sqrt{\frac{1-x}{1+x}} - \frac{\sqrt{\frac{1-x}{1+x}}}{1+\frac{1-x}{1+x}} + C \\ &= x \tan^{-1}\sqrt{\frac{1-x}{1+x}} - \frac{\sqrt{1-x^2}}{2} + C \end{align*}$$

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• $(1)$ : substitute $y=\sqrt{\dfrac{1-x}{1+x}}$
• $(2)$ : integrate by parts