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Integral((tan^4x)/4)dx?

5 Answers

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  • PaulR2
    Lv 7
    7 years ago
    Favorite Answer

    ∫ ( tan⁴x / 4 ) dx

    Remember, ∫ tan^n(x) dx = tan^(n-1)(x) - ∫ tan^(n-2)(x) dx

    Ignore the ¼, it's a constant we'll just divide the final answer by 4.

    ∫ ( tan⁴x ) dx = ( tan³x / 3 ) - ∫ tan²x dx

    Rememeber, tan²x + 1 = sec²x

    ∫ ( tan⁴x ) dx = ( tan³x / 3 ) - ∫ ( sec²x - 1 ) dx

    ∫ ( tan⁴x ) dx = ( tan³x / 3 ) - ( tan(x) - x )

    ∫ ( tan⁴x ) dx = ( tan³x / 3 ) - tan(x) + x

    ∫ ( tan⁴x / 4 ) dx = ( ( tan³x / 3 ) - tan(x) + x ) / 4

  • 7 years ago

    int tan^4x/4 dx

    =1/4 int tan^2x* tan^2x dx

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

    = 1/4int tan^2x* sec^2x dx - 1/4 int tan^2x dx

    = 1/4 tan^3x/3 - 1/4[ int (sec^2x - 1) dx

    = 1/12 tna^3x - 1/4[tanx + x/4 +C

    = 1/12[ 3x +tanx( tan^2x- 3) + C

    = 1/12 [ 3x +tanx( sec^2x - 4) ] +c

  • Anonymous
    7 years ago

    tan^(4x)/4 = [ tan^2(x) (1 +tan^2(x) ) - (tan^2 (x) + 1) + 1]/4

    integral of [ tan^2(x) (1 +tan^2(x) ) - (tan^2 (x) + 1) + 1]/4 dx = (1/4)[ tan^3(x) / 3 - tan(x) + x] + C

  • 7 years ago

    take

    I =Integral tan^4x dx

    I = Integral tan^2x.tan^2x dx

    I = integral tan^2x(sec^2x-1) dx

    I = Integra [tan^2xsec^2x-tan^2x] dx

    I = Integraltan^2x.sec^2x dx - Integral tan^2x dx

    I = Integral tan^2x.d(tanx) - Integral tan^2x dx [ Integration by parts method]

    I = tan^2x.tanx - Integral tanx d(tan^2x) - Integral tan^2x dx

    I = tan^3x - Integral (tanx.2tanx sec^2x) dx - Integral tan^2x dx

    I = tan^3x - 2.Integral (tan^2x(1+tan^2x) dx - Integral tan^2x dx

    I = tan^3x - 2.integral[tan^2x+tan^4x] dx - Integral tan^2x dx

    I = tan^3x - 2.Integral tan^2x dx - 2.Integral tan^4x dx - Integral tan^2x dx

    I = tan^3x - 2I - 3.Integral tan^2x dx

    I+2I = tan^3x - 3.Integral (sec^2x-1) dx

    3I = tan^3x - 3 tanx -3x + C/3

    I = 1/3. tan^3x - tanx -x + C

    Integral tan^4x dx = 1/3. tan^3x - tanx -x + C

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  • 7 years ago

    int(tan^4x/4)dx = 1/4 int (tan²x tan²x)dx

    = 1/4 int [tan²x(sec²x -1)]dx

    = 1/4 int tan²xsec²x dx - 1/4 int(tan²x) dx

    = 1/4 int tan²xsec²x dx - 1/4 int(sec²x - 1) dx

    = 1/4 int tan²xsec²x dx - 1/4 int(sec²x)dx +1/4 int dx

    in the first term,

    u = tan x; du = sec²x; n = 2

    = (1/4)*[(tan^3x)/3] - 1/4 tanx + x/4 + C

    = 1/12 tan^3x - 1/4 tanx + x/4 + C

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