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Math Help - substitution rule

  1. #1
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    substitution rule

    Can someone check if these are correct. If they're not, please help me how to solve it the right way. Thanks a lot!!!
    1. integral e^cost sint dt
    u=cost d/du=-sint du=-sintdt , -du=sint dt
    =integral e^u -du= -integral e^cost + c
    2. integral sin(ln x)/x dx
    u=ln x d/du=1/x du=1/xdx, -1/xdu=dx
    =-1/x integral u du = -1/x(lnx)+c
    3. integral (1 + tan thelta)^5 sec^2 thelta d thelta
    u= 1+ tan thelta, d/du=-ln(cosx) du=-ln(cosx)dx, -du=ln(cos)xdx
    = integral u^5 (-du) = -1/6(1 + tan thelta)^6 +C
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  2. #2
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    Quote Originally Posted by jsu03 View Post
    Can someone check if these are correct. If they're not, please help me how to solve it the right way. Thanks a lot!!!
    1. integral e^cost sint dt
    u=cost d/du=-sint du=-sintdt , -du=sint dt
    =integral e^u -du= -integral e^cost + c
    2. integral sin(ln x)/x dx
    u=ln x d/du=1/x du=1/xdx, -1/xdu=dx
    =-1/x integral u du = -1/x(lnx)+c
    3. integral (1 + tan thelta)^5 sec^2 thelta d thelta
    u= 1+ tan thelta, d/du=-ln(cosx) du=-ln(cosx)dx, -du=ln(cos)xdx
    = integral u^5 (-du) = -1/6(1 + tan thelta)^6 +C
    It's so much easier to read if you use LaTeX...

    1. \int{e^{\cos{t}}\sin{t}\,dt}

    Let u = \cos{t} so \frac{du}{dt} = -\sin{t}

    \int{e^{\cos{t}}\sin{t}\,dt} = -\int{-e^{\cos{t}}\sin{t}\,dt}

     = -\int{e^u \frac{du}{dt}\, dt}

     = -\int{e^u \, du}

     = -e^u + C

     = -e^{\cos{t}} + C.
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  3. #3
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    Quote Originally Posted by jsu03 View Post
    Can someone check if these are correct. If they're not, please help me how to solve it the right way. Thanks a lot!!!
    1. integral e^cost sint dt
    u=cost d/du=-sint du=-sintdt , -du=sint dt
    =integral e^u -du= -integral e^cost + c
    2. integral sin(ln x)/x dx
    u=ln x d/du=1/x du=1/xdx, -1/xdu=dx
    =-1/x integral u du = -1/x(lnx)+c
    3. integral (1 + tan thelta)^5 sec^2 thelta d thelta
    u= 1+ tan thelta, d/du=-ln(cosx) du=-ln(cosx)dx, -du=ln(cos)xdx
    = integral u^5 (-du) = -1/6(1 + tan thelta)^6 +C
    2. \int{\frac{\sin{\ln{x}}}{x}\, dx}

    Let  u = \ln{x} so \frac{du}{dx} = \frac{1}{x}.

    So \int{\frac{\sin{\ln{x}}}{x}\, dx} = \int{\sin{u} \frac{du}{dx}\, dx}

     = \int{\sin{u} \, du}

     = -\cos{u} + C

     = -\cos{\ln{x}} + C.
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  4. #4
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    Quote Originally Posted by jsu03 View Post
    Can someone check if these are correct. If they're not, please help me how to solve it the right way. Thanks a lot!!!
    1. integral e^cost sint dt
    u=cost d/du=-sint du=-sintdt , -du=sint dt
    =integral e^u -du= -integral e^cost + c
    2. integral sin(ln x)/x dx
    u=ln x d/du=1/x du=1/xdx, -1/xdu=dx
    =-1/x integral u du = -1/x(lnx)+c
    3. integral (1 + tan thelta)^5 sec^2 thelta d thelta
    u= 1+ tan thelta, d/du=-ln(cosx) du=-ln(cosx)dx, -du=ln(cos)xdx
    = integral u^5 (-du) = -1/6(1 + tan thelta)^6 +C
    3. \int{(1 + \tan{\theta})^5 \sec^2{\theta}\, d\theta}

    Let u = 1 + \tan{\theta} so \frac{du}{d\theta} = \sec^2{\theta}.

    So \int{(1 + \tan{\theta})^5 \sec^2{\theta}\, d\theta} = \int{u^5 \frac{du}{d\theta}\, d\theta}

     = \int{u^5 \, d\theta}

     = \frac{1}{6}u^6 + C

     = \frac{1}{6}(1 + \tan{\theta})^6 + C.
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