# Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$

$$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$ I've tried substituting $x=2\tan^2y$, and I've got: $$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$ But I'm not entirely sure this is a good thing as I've been unable to proceed any further from there.

• I think integration by parts with u being the arctangent will work – Triatticus Nov 28 '15 at 21:07

Let $x=2 u^2$:

$$2 \sqrt{2} \int du \, \frac{u}{\sqrt{1+u^2}} \arctan{u} = 2\sqrt{2} \sqrt{1+u^2} \arctan{u} - 2 \sqrt{2} \int \frac{du}{\sqrt{1+u^2}}$$

The latter integral is easily done using the sub $u=\sinh{v}$, so we have as the integral

$$2 \sqrt{x+2} \arctan{\sqrt{\frac{x}{2}}} - 2 \sqrt{2} \log{\left (\sqrt{\frac{x}{2}}+\sqrt{1+\frac{x}{2}} \right ) }+C$$

I've got $~\dfrac1{\sqrt2}\displaystyle\int\frac{y\sin y}{\cos^4 y}~dy,~$ but I'm not entirely sure this is a good thing.

Of course it is ! Integrate by parts with regard to $f'(y)=\dfrac{\sin y}{\cos^4y}=-\dfrac{\cos'y}{\cos^4y}=\bigg(\dfrac1{3\cos^3y}\bigg)'.$

Then rewrite $~\dfrac1{\cos^3y}~$ as $~\dfrac{\cos y}{\cos^4y}=\dfrac{\sin'y}{\big(1-\sin^2y\big)^2}~,~$ and use an appropriate substitution to

reduce the integrand to a rational fraction.

Let $\alpha=\arctan\sqrt{\frac{x}{2}}$

$$I=\int \frac{\alpha dx}{\sqrt{x+2}}=\int\alpha d(2\sqrt{x+2})=2\alpha\sqrt{x+2}-2\int \sqrt{x+2}\space d\alpha$$

The calculation gives $$d\alpha=\frac{\sqrt {2} dx}{\sqrt x(x+2)}$$

Hence $I=2\alpha\sqrt{x+2}-2\sqrt 2\int \frac{\sqrt{x+2}\space dx}{(x+2)\sqrt x}$

$I=2\alpha\sqrt{x+2}-2\sqrt 2\int\frac{dx}{\sqrt{x(x+2)}}$

Now $\int \frac{dx}{\sqrt{x(x+2)}}=2$ $\sin h^{-1}{\sqrt\frac{x}{2}}=2\ln(\sqrt {\frac{x}{2}}+ \sqrt{\frac{x}{2}+1})$

Thus $$I=2\alpha\sqrt{x+2}-4\sqrt2\ln \left(\sqrt {\frac{x}{2}}+ \sqrt{\frac{x}{2}+1}\right)+ Constant$$

Notice, $$\int \frac{\tan^{-1}\sqrt{\frac{x}{2}}}{\sqrt{x+2}}\ dx=\int \frac{\tan^{-1}\sqrt{\frac{x}{2}}}{\sqrt 2\sqrt{\frac{x}{2}+1}}\ dx$$ now, let $\frac{x}{2}=\tan^2\theta\implies dx=2\tan\theta\sec^2\theta \ d\theta$, $0\le \theta\le \pi/2$ $$=\frac{1}{\sqrt2}\int \frac{\tan^{-1}\left(\tan\theta\right)}{\sqrt{\tan^2\theta+1}}(2\tan\theta\sec^2\theta \ d\theta)$$ $$=\frac{2}{\sqrt2}\int \frac{\theta\sec^2\theta\tan\theta}{\sec\theta}\ d\theta$$ $$=\sqrt 2\int \theta(\sec\theta\tan\theta)\ d\theta$$ $$=\sqrt 2\left(\theta\sec\theta-\int \sec\theta \ d\theta\right)$$ $$=\sqrt 2\left(\theta\sec\theta-\ln\left|\sec\theta+\tan\theta\right|\right)+C$$

$$=2\sqrt{x+2} \tan^{-1} \sqrt{\frac x 2} - 4\sqrt{x+2} + C$$

• yes, you are right, we are considering positive values only. – Harish Chandra Rajpoot Nov 28 '15 at 21:27
• for ease of calculation. Although, I have put assumption $0\le \theta\le \pi/2$ – Harish Chandra Rajpoot Nov 28 '15 at 21:30

\begin{align} u & = \arctan \sqrt{\frac x 2} \\[10pt] du & = \frac{dx/2}{\left(1+ \dfrac x 2\right)2\sqrt{\dfrac x 2}} = \frac{dx}{(2+x)\sqrt{2x}} \\[10pt] dv & = \frac{dx}{\sqrt{x+2}} \\[10pt] v & = 2\sqrt{x+2} \end{align} \begin{align} \int u\,dv & = uv - \int v\,du = 2\sqrt{x+2} \arctan \sqrt{\frac x 2} - \int \frac{\sqrt 2\,dx}{\sqrt{x+2}\sqrt x} \end{align}

Then $$\sqrt{x^2+2x} = \sqrt{(x^2+2x + 1) - 1} = \sqrt{(x+1)^2 -1} = \sqrt{\sec^2\theta - 1},\quad\text{etc.}$$

the taylor series of $$\frac{\sin y}{\cos^4y}=\sum_{n=1}^{\infty }\frac{(2(2n-1)!-1)y^{2n-1}}{(2n-1)!}$$ so $$\frac{1}{\sqrt2}\int\frac{y\sin y}{cos^4 y}dy=\frac{1}{\sqrt2}\int\sum_{n=1}^{\infty }\frac{(2(2n-1)!-1)y^{2n}}{(2n-1)!}dy$$ $$=\frac{1}{\sqrt2}\sum_{n=1}^{\infty }\frac{(2(2n-1)!-1)y^{2n+1}}{(2n+1)(2n-1)!}+C$$