Question

expansion of tan inverse x

Answer 1

1/(1+x^2)..........................

Answer 2

Answer:

$tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+........................$

Step-by-step explanation:

Expansion of tan inverse x

I have found expansion of tan inverse x by integration method.

we know that the expansion of $\frac{1}{1+x}$

$\frac{1}{1+x}=1-x+x^2-x^3+..............$

Replace x by x², we get

$\frac{1}{1+x^2}=1-x^2+x^4-x^6+..............$

Integrating on both sides we get,

$\int{\frac{1}{1+x^2}}\:dx=\int{[1-x^2+x^4-x^6+..............}]\:dx$

$tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+..............+C$

Applying the condition,

$tan^{-1}0=0$ we get, C=0

Therefore

$tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+........................$