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# integral help! sec^4x dx???!!?

integral from (PI/3) to (2pi/3) of sec^4x dx

so far I have

integral from (PI/3) to (2pi/3) of (1+tan^2x)sec^2x dx

### 3 Answers

- cidyahLv 76 years agoFavorite Answer
∫ sec^4 x dx = ∫ sec^2 x sec^2 x dx

Integrate by parts

dv= sec^2 x dx; v = tan x;

u = sec^2 x ; du = 2 sec^2 x tan x

∫ u dv = u v - ∫ v du

∫ sec^4 x dx = tan x sec^2 x - 2 ∫ sec^2 x tan^2 x dx

∫ sec^4 x dx = tan x sec^2 x - 2 ∫ sec^2 x (sec^2 x-1) dx

∫ sec^4 x dx = tan x sec^2 x - 2 ∫ sec^4 x + 2 ∫ sec^2 x dx

3 ∫ sec^4 x dx = tan x sec^2 x + 2 ∫ sec^2 x dx

3 ∫ sec^4 x dx = tan x sec^2 x + 2 tan x

∫ sec^4 x dx = (1/3) tan x sec^2 x + (2/3) tan x + C

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- PolyhymnioLv 76 years ago
∫sec⁴x dx = ∫sec²x sec²x dx = ∫(1 + tan²x)sec²x dx

Let u = tanx, du = sec²x dx

∫(1 + u²) du = u + u³/3 + C =

tanx + tan³x/3 + C

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- DaveLv 66 years ago
You were going the right way, now use u=tan(x) and du=sec^2(x)dx

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