Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

Can you help me find the antiderivative of...? please...help?

f(x) = (sinx + cosx)^2

Im reviewing for a test and came across this question and just cant seem to get it!...help would be greatly appreciated...

4 Answers

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  • 9 years ago
    Favorite Answer

    I think f(x) should be f'(x) in your question.

    i.e. f'(x) = (sinx +cosx)^2

    = sinx^2 +cosx^2 +2sinxcosx

    = 1+ 2sinxcosx

    = 1 + sin2x

    Integrating

    f(x) = x + 2cos2x+C

    Source(s): Me
  • 9 years ago

    Multiply it all out:

    sin^2x + 2sinxcosx + cos^2x dx

    Rearrange:

    sin^2x + cos^2x + 2sinxcosx dx

    1 + 2sinxcosx dx

    Break it up into two integrals:

    Unfortunately this one needs a u sub

    ∫1 + ∫2sinxcosx dx

    Only do u sub for right side:

    u = sinx

    du = cosx dx

    ∫1 + ∫2 u du -----> the u we said is sinx, and the "du" sucked up the cosxdx

    Now we have:

    x + 2((u^2)/2) + C

    = x + u^2 + C

    Replace u with what we said it was..... remember we said u= sinx

    x + sinx^2 + C

  • ?
    Lv 4
    9 years ago

    f(x) = (sinx + cosx)^2

    f(x) = sin^2(x) + 2sinxcosx + cos^2(x)

    *sin^2(x) + cos^2(x) = 1

    f(x) = 2sinxcosx + 1

    *2sinxcosx = sin(2x)*

    f(x) = 1 + sin(2x)

    int(1+sin(2x)) = x + -(1/2)cos(2x) + C

    antiderivative of f(x) = x - (1/2)cos(2x) + C

    Without being arrogant, I'll just tell you that I am the only person who has yet given you the correct answer.

    And to Sydney, if you integrate a derivative your just get the original function. There is no reason you can only find the derivative of f'(x)...

  • 9 years ago

    you would to take the integral of that which is integral ((sin(x)+cos(x))^2 dx

    = integral (sin^2(x) + 2sin(x)*cos(x)+ cos^2(x)) dx

    = -(1/2)cos(2x)+x

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