leprofece said:integral of arctg(x)/sqrt(1-x^2)
maybe u = arctg x du = 1 /1+x^2
but x = tg u
maybe this is the way isn't it?
chisigma said:A possible approach is to use the MacLaurin expansion...
$\displaystyle \frac{\tan^{-1} x}{\sqrt{1-x^{2}}} = x + \frac{1}{6}\ x^{3} + \frac{49}{120}\ x^{5} + \frac{81}{560}\ x^{7} + \mathcal{O}\ (x^{9})\ (1)$
... and to integrate term by term...
Kind regards $\chi$ $\sigma$
leprofece said:Thank for your answer but you must solve it by normal or traditional methods students have not studied series yet
Could anybody do that way??
The formula for solving integrals using integration by parts is ∫u dv = uv - ∫v du, where u and v are functions of x and dv and du are their respective differentials.
When using integration by parts, the mnemonic "LIATE" (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) can be helpful in determining which function to choose as u. The function that comes first in the order should be chosen as u.
Integration by parts can be used on any integral, but it may not always lead to a simpler solution. It is best used when one function is easy to integrate and the other is easy to differentiate.
When solving integrals involving arctangent and square root functions, it is helpful to use the substitution method to simplify the integral before applying integration by parts. For example, substituting x = sinθ can simplify integrals involving arctangent.
There are no specific rules for choosing u and v when solving integrals involving arctangent and square root functions. It may require some trial and error to determine the best choice for u and v, but generally, choosing the more complicated function as u can lead to a simpler solution.