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Thread: Problem with Integration of trig functions

  1. #1
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    Problem with Integration of trig functions

    $\displaystyle \int\frac{2cos^2(2t)-1}{1+3sin4t}dx$


    $\displaystyle \int\frac{cot^2ln(1-x)}{1-x}dx$

    $\displaystyle \int\frac{tan^2e^-x}{e^x}dx$


    I am having problems with simplifying the given, before I can Integrate.
    For example, the problem 1, my u is the denominator, but after deriving the u It is still impossible to solve it, same problem applies to problem 2. For number I just can't solve it

    any help is appreciated, thank you
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  2. #2
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    Quote Originally Posted by honestliar View Post
    $\displaystyle \int\frac{2cos^2(2t)-1}{1+3sin4t}dx$


    $\displaystyle \int\frac{cot^2ln(1-x)}{1-x}dx$

    $\displaystyle \int\frac{tan^2e^-x}{e^x}dx$


    I am having problems with simplifying the given, before I can Integrate.
    For example, the problem 1, my u is the denominator, but after deriving the u It is still impossible to solve it, same problem applies to problem 2. For number I just can't solve it

    any help is appreciated, thank you
    1. note that $\displaystyle 2\cos^2(2t) - 1 = \cos(4t)$ , let $\displaystyle u = 1+3\sin(4t)$

    2. use the identity $\displaystyle \cot^2{t} = \csc^2{t} - 1$ , let $\displaystyle u = \ln(1-x)$

    3. use the identity $\displaystyle \tan^2{t} = \sec^2{t} - 1$ , let $\displaystyle u = e^{-x}$
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