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Math Help - Evaluating integrals

  1. #1
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    Evaluating integrals

    Can someone please help me to evaluate these integrals
    sin ^5 x dx
    tan 3 x dx
    Do I simply apply rule one for adding the the exponent and placing the same exponent in the denominator? Do I substitue -cos for sin? Do I substitute sec^2 x for tan x? Help please.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by bosmith View Post
    Can someone please help me to evaluate these integrals
    sin ^5 x dx
    tan 3 x dx
    Do I simply apply rule one for adding the the exponent and placing the same exponent in the denominator? Do I substitue -cos for sin? Do I substitute sec^2 x for tan x? Help please.
    Note: I'll be using the fact that \sin^2(x)=1-\cos^2(x) a lot without explicitly stating it.

    \int\sin^5(x)\,dx = \underbrace{\int\sin^3(x)\,dx}_{1st~integral}-\underbrace{\int\sin^3(x)\cos^2(x)\,dx}_{2nd~integ  ral}

    First Integral: \int\sin^3(x) = \int\sin(x)-\sin(x)\cos^2(x)\,dx

    Let u=\cos(x) so du = -\sin(x)\,dx. We have: -\int\,du +\int u^2\,du = -u+\frac{u^3}{3} = -\cos(x)+\frac{\cos^3(x)}{3}

    Second Integral: \int\sin^3(x)\cos^2(x)\,dx = \int\sin(x)\cos^2(x)-\sin(x)\cos^4(x)\,dx

    Let u=\cos(x) so du = -\sin(x)\,dx. We have:

    -\int u^2\,du +\int u^4\,du = -\frac{u^3}{3}+\frac{u^5}{5} = -\frac{\cos^3(x)}{3}+\frac{\cos^5(x)}{5}

    -\cos(x)+\frac{\cos^3(x)}{3} - \left(-\frac{\cos^3(x)}{3}+\frac{\cos^5(x)}{5}\right) = \boxed{-\frac{1}{5}\cos^5(x) + \frac{2}{3}\cos^3(x)-\cos(x)}
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by bosmith View Post
    Can someone please help me to evaluate these integrals
    sin ^5 x dx
    tan 3 x dx
    Do I simply apply rule one for adding the the exponent and placing the same exponent in the denominator? Do I substitue -cos for sin? Do I substitute sec^2 x for tan x? Help please.
    \int\tan(3x)\,dx = \int\frac{\sin(3x)}{\cos(3x)}\,dx

    Let u=\cos(3x) and du=-3\sin(3x)\,dx. We now have:

    -\frac{1}{3}\int\frac{1}{u}\,du = -\frac{1}{3}\ln|u| = \boxed{-\frac{1}{3}\ln|\cos(3x)|}

    Or did you mean \int\tan^3(x)\,dx ?
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  4. #4
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    redsoxfan325

    Thanks for showing me that I was on the wrong track.
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  5. #5
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    Quote Originally Posted by bosmith View Post
    Can someone please help me to evaluate these integrals
    sin ^5 x dx
    tan 3 x dx
    Do I simply apply rule one for adding the the exponent and placing the same exponent in the denominator? Do I substitue -cos for sin? Do I substitute sec^2 x for tan x? Help please.
    I'm not sure what you mean by "adding the exponent". If you were thinking of the power rule, no, because this is x^n dx, you cannot do that.

    Because of the odd power in sin^5 t, you can factor out one "sin t" to use with the derivative: \int sin^5 x dx= \int sin^4 x (sin x dx) = (1- cos^2 x)^2 (sin x dx). Now use the substitution u= cos x, du= -sin x dx so the integral becomes -\int (1- u^2)^2 du. Multiply that out and you can apply the power rule to each part.

    For \int tan^3 x dx you can't "substitute sec^2 x for tan x"- they are not equal! It is the derivative of tan x that is equal to sec^2 x and you don't have any derivative of tan x in this. I would write it in the form \int \frac{sin^2 x}{cos^3 x} dx. Again, that is an odd power of sine so we can factor out one to use with dx:
    \int \frac{sin^2 x}{cos^3 x}(sin x dx). Again, let u= cos(x) so that du= -sin x dx so the integral becomes -\int \frac{1- cos^2 x}{cos^3 x} (cos x dx)= -\int \frac{1- u^2}{u^3}du.
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  6. #6
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    hallsofivy

    Thank you. I am struggling trying to lean Calculus. I was on the wrong track (confused). I now have a better understanding of the substitution fule, usign u= cosx. I will keep trying. I guess I am just not that smart. I have to leanr to ask questions sooner or somehow try to comprehend faster. I have a test in a couple of hours. Too bad for me that nobody seems to think it beneficial just to show me how to do the problems. But I truly thank you for putting me on the right track.
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