Find the indefinite integral [(cosx)^6 - (sinx)^6]dx?

Find the indefinite integral [(cosx)^6 - (sinx)^6]dx

I realize that there's power reduction formulas out there that I can remember, but in all honesty, I couldn't remember the power reduction formula for cosx or sinx raised to the 6th power if my life depended on it. Is there a systematic way of doing this problem?

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  • kb
    Lv 7
    9 years ago
    Best Answer

    Note that with a mix of algebra and trig identities:

    cos^6(x) - sin^6(x)

    = (cos^3(x) + sin^3(x)) (cos^3(x) - sin^3(x))

    = (cos x + sin x) (cos^2(x) - cos x sin x+ sin^2(x)) * (cos x - sin x) (cos^2(x) + cos x sin x+ sin^2(x))

    = (cos x + sin x) (1 - cos x sin x) * (cos x - sin x) (1 + cos x sin x)

    = (1 - cos^2(x) sin^2(x)) (cos^2(x) - sin^2(x)), rearranging the factors

    = (1 - (1/4) (2 sin x cos x)^2) cos(2x)

    = [1 - (1/4) sin^2(2x)] cos(2x).

    So, ∫ (cos^6(x) - sin^6(x)) dx

    = ∫ [1 - (1/4) sin^2(2x)] cos(2x) dx

    = ∫ [1 - (1/4) z^2] (dz/2), letting z = sin(2x)

    = (1/2) (z - z^3/12) + C

    = (1/24) (12z - z^3) + C

    = (1/24) (12 sin(2x) - sin^3(2x)) + C.

    I hope this helps!

  • mic
    Lv 4
    3 years ago

    nicely, it extremely is purely -cos(?x) + C. you comprehend that d(sin x)/dx = cos(x) and d(cos x)/dx = -sin(x). So it stands to reason that d(-cos x)/dx = -sin(x), or, in different words, ?sin(x)dx = -cos(x). And the ? purely pops in and out as a results of chain rule.

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