What's the integral of $\int {(tan x)}^{2} + {(tan x)}^{4} d x$ $int (tanx)^2+(tanx)^4 dx$ ?

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Answer 1 · 2016-04-27T03:27:30

$\frac{{tan}^{3} x}{3} + C$ $tan^3x/3+C$

Factor a ${tan}^{2} x$ $tan^2x$ inside the integral.

$= \int {tan}^{2} x (1 + {tan}^{2} x) d x$ $=inttan^2x(1+tan^2x)dx$

We know $1 + {tan}^{2} x = {sec}^{2} x$ $1+tan^2x=sec^2x$ through the Pythagorean identity.

$= \int {tan}^{2} x {sec}^{2} x d x$ $=inttan^2xsec^2xdx$

We can now use substitution:

$u = tan x \Rightarrow d u = {sec}^{2} x d x$ $u=tanx" "=>" "du=sec^2xdx$

This gives us:

$= \int u^{2} d u = \frac{u^{3}}{3} + C = \frac{{tan}^{3} x}{3} + C$ $=intu^2du=u^3/3+C=tan^3x/3+C$

What's the integral of ∫(tanx)2+(tanx)4dxint (tanx)^2+(tanx)^4 dx?