Answer:
See below
Step-by-step explanation:
[tex]\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x[/tex]
Recall
[tex]\dfrac{d}{dx}\tan x=\sec^2[/tex]
Using the chain rule
[tex]\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}[/tex]
such that [tex]u = \tan x[/tex]
we can get a general formulation for
[tex]y = \tan^n x[/tex]
Considering the power rule
[tex]\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}[/tex]
we have
[tex]\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x[/tex]
therefore,
[tex]\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x[/tex]
Now, once
[tex]\sec^2 x - 1= \tan^2x[/tex]
we have
[tex]3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x[/tex]
Hence, we showed
[tex]\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x[/tex]
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For the integration,
[tex]$\int \sec^4 x\, dx $[/tex]
considering the previous part, we will use the identity
[tex]\boxed{\sec^2 x - 1= \tan^2x}[/tex]
thus
[tex]$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$[/tex]
and
[tex]$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx $[/tex]
Considering [tex]u = \tan x[/tex]
and then [tex]du=\sec^2x\ dx[/tex]
we have
[tex]$\int u^2 \, du = \dfrac{u^3}{3}+C$[/tex]
Therefore,
[tex]$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$[/tex]
[tex]$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$[/tex]
