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

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  • cidyah
    Lv 7
    6 years ago
    Favorite 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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  • 6 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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  • Dave
    Lv 6
    6 years ago

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

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