Question

Integrate the function w.r.t.x. : $\tan^{-1}( \frac{2\tan x}{1-\tan^{2}x})$

Answer 1

Answer:

$\bf\:\int{tan^{-1}(\frac{2tanx}{1-tan^2x})}\:dx=x^2+c$

Step-by-step explanation:

I have applied change of variable method to solve this integral

$\int{tan^{-1}(\frac{2tanx}{1-tan^2x})}\:dx$

Using, the identity

$\boxed{\bf\:tan2A=\frac{2\:tanA}{1-tan^2A}}$

$=\int{tan^{-1}(tan2x)}\:dx$

Using

$\boxed{tan^{-1}(tanx)=x}$

$=\int{2x}\:dx$

$=2\:\int{x}\:dx$

$=2[\frac{x^2}{2}]+c$

$=x^2+c$

$\bf\:\int{tan^{-1}(\frac{2tanx}{1-tan^2x})}\:dx=x^2+c$

Answer 2

Solution :

We know that,

tan2x = 2 tanx/(1 - tan²x)

Now, ∫ tan⁻¹ {2 tanx/(1 - tan²x)} dx

= ∫ tan⁻¹ (tan2x) dx

= ∫ 2x dx [ ∵ tan⁻¹ (tanmx) = mx ]

= 2 ∫ x dx

= 2 (x²/2) + c

where c is integral constant

= x² + c ,

which is the required integral.