I was trying to solve $\int \frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2} dx$. I think I have the first step right, which is $\int \sqrt{\arctan\frac{1}{x}} d(\arctan x) $, but I'm not sure how to proceed. I considered integration by parts, but it seems to make the new integrand too complicated. please help. Thanks.
Asked
Active
Viewed 170 times
1
-
Try $u=arctan(1/x)$. – Paul Apr 03 '16 at 00:25
-
Go to http://www.integral-calculator.com/ and type this in: ((arctan(1/x))^(1/2))/(1+x^2) . It shows the steps and the solution – Apr 03 '16 at 01:12
3 Answers
2
$$ u = \arctan\left(\frac{1}{x}\right) \rightarrow du = \frac{1}{1+\frac{1}{x^2}} \cdot \frac{-1}{x^2} dx = \frac{-1}{1+x^2} dx $$ $$ \int \frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2} dx = \int -\sqrt{u} du = -\frac{2}{3}u^{3/2} + C = -\frac{2}{3}\left(\arctan\left(\frac{1}{x}\right)\right)^{3/2} + C $$ So: $$ \int \frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2} dx = -\frac{2}{3}\left(\arctan\left(\frac{1}{x}\right)\right)^{3/2} + C $$
Anonymous
- 1,158
0
Sunstitute $u/\tan^{-1}(1/x)$. Thus $x=\cot(u), dx=-\csc^2(u) du$. You should find the trigonometric functions cancel out and the integral is then easy.
Oscar Lanzi
- 39,403
0
HINT:
Using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?, as $x>0,$ $$\arctan\dfrac1x=\text{arccot}x=\dfrac\pi2-\arctan x$$
Let $\arctan x=u$
lab bhattacharjee
- 274,582