Question

tan⁴x.dx
Integrate the function

Answer 1

$\huge{Be Brainly...}$

Answer 2

Answer:

$=\dfrac{tan^{3} x}{3} - tanx +x +c$

Step-by-step explanation:

Here, the given equation is

$=\int tan^{4}x  dx$

and we find the integration of the given function, first we break it into two parts, and then put the value of $tan^{2}x=sec^{2}x-1$ in one part, and then integrate it

$=\int tan^{2}x tan^{2}xdx\\=tan^{2}x=sec^{2}x-1\\=\int tan^{2}x(sec^{2}x-1)dx\\=\int tan^{2}xsec^{2}xdx-\int tan^{2}xdx\\let \\tan x =t\\sec^{2}xdx=dt\\=\int t^{2} dx - \int (sec^{2}x-1)dx\\=\int t^{2} dx - \int sec^{2} xdx + \int dx\\\\=\dfrac{t^{3} }{3} - tanx +x +c\\\\=\dfrac{tan^{3}x }{3} - tanx +x +c$