What is the integral of ${(cos x + sec x)}^{2}$ $( cos x +sec x )^2$ ?

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What is the integral of ${(cos x + sec x)}^{2}$ $( cos x +sec x )^2$ ?

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Answer 1 · 2016-08-01T17:41:52

$\frac{1}{4} sin (2 x) + \frac{5}{2} x + tan x + C$ $1/4sin(2x)+5/2x+tanx+C$

Explanation:

Note that ${(cos x + sec x)}^{2} = {cos}^{2} x + 2 sec x cos x + {sec}^{2} x = {cos}^{2} x + 2 + {sec}^{2} x$ $(cosx+secx)^2=cos^2x+2secxcosx+sec^2x=cos^2x+2+sec^2x$ .

Thus, we have:

$\int {cos}^{2} x d x + 2 \int d x + \int {sec}^{2} d x$ $intcos^2xdx+2intdx+intsec^2dx$

The last two are common integrals:

$= \int {cos}^{2} x d x + 2 x + tan x$ $=intcos^2xdx+2x+tanx$

For the remaining integral, use the following identity:

$cos (2 x) = 2 {cos}^{2} x - 1 \Rightarrow {cos}^{2} x = \frac{cos (2 x) + 1}{2}$ $cos(2x)=2cos^2x-1" "=>" "cos^2x=(cos(2x)+1)/2$

Thus the integral equals:

$= \frac{1}{2} \int cos (2 x) d x + \frac{1}{2} \int d x + 2 x + tan x$ $=1/2intcos(2x)dx+1/2intdx+2x+tanx$

$= \frac{1}{2} \int cos (2 x) d x + \frac{1}{2} x + 2 x + tan x$ $=1/2intcos(2x)dx+1/2x+2x+tanx$

$= \frac{1}{2} \int cos (2 x) d x + \frac{5}{2} x + tan x$ $=1/2intcos(2x)dx+5/2x+tanx$

For the first integral, let $u = 2 x$ $u=2x$ , so $d u = 2 d x$ $du=2dx$ .

$= \frac{1}{4} \int 2 cos (2 x) d x + \frac{5}{2} x + tan x$ $=1/4int2cos(2x)dx+5/2x+tanx$

$= \frac{1}{4} \int cos (u) d u + \frac{5}{2} x + tan x$ $=1/4intcos(u)du+5/2x+tanx$

$= \frac{1}{4} sin (u) + \frac{5}{2} x + tan x + C$ $=1/4sin(u)+5/2x+tanx+C$

$= \frac{1}{4} sin (2 x) + \frac{5}{2} x + tan x + C$ $=1/4sin(2x)+5/2x+tanx+C$

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What is the integral of ${(cos x + sec x)}^{2}$ $( cos x +sec x )^2$ ?

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Explanation:

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