Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Evaluate the indefinite integral: ∫ cos(θ) sin^6(θ) d(θ)?

Please help! I don't understand how to solve this.

7 Answers

  • 1 decade ago
    Best Answer

    I'll use the | symbol to represent the integral symbol.

    I'll use x for theta

    Let's do a u-substitution:

    u = sin(x)

    du = cos(x) dx

    dx = du/cosx

    |cosx(u^6)du/cosx the cosx cancels out

    |u^6 du

    (1/7)u^7 + C replace u with sinx

    (1/7)(sin^7x) + C


  • Anonymous
    1 decade ago

    ok, the method that u are going to use for this problem is called a u substitution.

    ∫ cos(θ) sin^6(θ) d(θ)

    first let u = sin (θ)

    implicitly differentiate u to with respect to θ

    d/dθ [ u = sin (θ) ]

    du / dθ = cos (θ)

    now cross multiply.

    du = cos (θ) dθ

    so u = sin (θ) and du = cos (θ) dθ. plug in u for sin (θ) and du for cos (θ) dθ. notice that cos (θ) dθ is conviently a part of the integrand

    ∫ cos (θ) sin^6(θ) dθ

    = ∫ [ sin (θ) ]^6 [ cos (θ) dθ ]

    the sin (θ) inside the bracket is u. the cos (θ) dθ inside the other bracket is du

    = ∫ u^6 du

    = (u^7 / 7) + C

    now remember when we let u = sin θ? now we have to plug sin θ back in for u

    = (sin^7(θ) / 7) + C

    and theres the answer. the hardest part of a u substitution is figuring out which substitution to make. wen u get really good at u substitutions u can do the problem in your head. like before i even did this problem, i already knew the answer was going to be sin^7(θ) / 7 + C.

  • Anonymous
    1 decade ago

    You substitute a variable for the trig function that has the higher power usually. That is the common method in dealing with those integrals

    z = sin(t)

    dz = cos(t) dt

    INT z^6 dz = z^7/ 7 = sin^7(t) /7 + C

  • Hello ,

    If u look well , u can realize that ∫sin(θ) is already existed but without


    so :

    ∫ cos(θ) sin^6(θ) d(θ) = - ∫ sin^6(θ) *(-cos(θ) d(θ)) = - (1/7)sin^7(θ)+C

    Good Luck

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  • 3 years ago

    this indefinite fundamental could be solved utilising u substitution. i'm utilising x as theta reason i don't have that image set u=sinx then Du= cosx u get the fundamental of a million/u^2 Du then carry u to the coolest making it the fundamental of u^-2 Du utilising integration rules the fundamental is u^-a million/-a million then in basic terms substitute sinx for u and you get -(a million/sinx)

  • Como
    Lv 7
    1 decade ago

    Let u = sin Ө

    du = cos Ө dӨ

    I = ∫ u^6 du

    I = u^7 / 7 + C

    I = ( sin Ө )^7 / 7 + C

  • 1 decade ago

    ∫ cos(θ) sin^6(θ) d(θ)?

    let u = sin (θ)

    => du = cos(θ) d(θ)

    ∫ cos(θ) sin^6(θ) d(θ)

    = ∫ sin^6(θ) cos(θ)d(θ)

    = ∫ u^6 du

    = u^7 /7

    = sin^7(θ) / 7 + C


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