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Math Help - Integration by Parts

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    Integration by Parts

    I need help with 4 of these. I tried them, but I keep getting the wrong answer.

    1. Evalute the integral of (sin(8x))^3*(cos(8x))^2*dx. Use C to denote an arbitrary constant.

    2. Evalute the integral of (cos(5x))^3*dx. Use C to denote an arbitrary constant.

    3. Evalute the integral of csc(3x)dx. Use C to denote an arbitrary constant.

    4. Evalute the integral of (2-2sinx)/cosx dx. Use C to denote an arbitrary constant.
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    Quote Originally Posted by asnxbbyx113 View Post
    4. Evalute the integral of (2-2sinx)/cosx dx. Use C to denote an arbitrary constant.
    Hint:

    \int \frac{2-2\sin x}{\cos x} dx = \int \frac{2}{\cos x} - \frac{2\sin x}{\cos x} dx

    = \int 2\sec x - 2\tan x dx
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    Quote Originally Posted by asnxbbyx113 View Post

    3. Evalute the integral of csc(3x)dx. Use C to denote an arbitrary constant.
    .
     \int \csc 3x dx

    Let t=3x:

    \frac{1}{3} \int \csc t dt

    =\frac{1}{3} \left( \ln|\csc x - \cot x| + C\right)
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    Quote Originally Posted by asnxbbyx113 View Post
    I need help with 4 of these. I tried them, but I keep getting the wrong answer.

    1. Evalute the integral of (sin(8x))^3*(cos(8x))^2*dx. Use C to denote an arbitrary constant.

    2. Evalute the integral of (cos(5x))^3*dx. Use C to denote an arbitrary constant.
    .
    \int \cos^3 x dx = \int \cos x \cos^2 x dx = \int \cos x (1-\sin^2 x) dx

    Let t=\sin x then t'=\cos x:

    \int (1-t^2)dt = t - \frac{1}{3}t^3+C

    \sin x - \frac{1}{3}\sin^3 x +C

    Now to find,

    \int \cos^3 5x dx

    Just use the substitution u=5x und do what I do above.
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    Quote Originally Posted by asnxbbyx113 View Post
    I need help with 4 of these. I tried them, but I keep getting the wrong answer.

    1. Evalute the integral of (sin(8x))^3*(cos(8x))^2*dx. Use C to denote an arbitrary constant.
    \int \sin^3 8x \cos^2 8x dx

    =\frac{1}{8} \int \sin^3 u \cos^2 u du

    =\frac{1}{8} \int \sin^2 u \cos^2 u \sin u du

    =\frac{1}{8} \int (1-\cos^2 u)\cos^2 u \sin u du

    Let t=\cos u then,

    -\frac{1}{8} \int (1-t^2)t^2 dt

    You finish!
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