What is the Integral of ${(1 + tan x)}^{5} \cdot ({sec}^{2} (x)) d x$ $(1+tanx)^5 * (sec^2(x)) dx$ ?

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Answer 1 · 2016-12-20T14:53:50

The answer is $= \frac{{(1 + tan x)}^{6}}{6} + C$ $=(1+tanx)^6/6+C$

We must calculate $\int {(1 + tan x)}^{5} {sec}^{2} x d x$ $int(1+tanx)^5sec^2xdx$

We do this integral by substitution

Let $u = 1 + tan x$ $u=1+tanx$

$d u = {sec}^{2} x d x$ $du=sec^2xdx$

$\int {(1 + tan x)}^{5} {sec}^{2} x d x$ $int(1+tanx)^5sec^2xdx$

$= \int u^{5} d u = \frac{u^{6}}{6}$ $=intu^5du=u^6/6$

$= \frac{{(1 + tan x)}^{6}}{6} + C$ $=(1+tanx)^6/6+C$

What is the Integral of (1+tanx)5⋅(sec2(x))dx (1+tanx)^5 * (sec^2(x)) dx ?