3
$\begingroup$

$$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$ 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.

$\endgroup$
  • $\begingroup$ I think integration by parts with u being the arctangent will work $\endgroup$ – Triatticus Nov 28 '15 at 21:07
4
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.