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    Find the integral of 1/(coshx+asinh(x))dx distinguishing between various possibilities of a. Not a clue how to start on this one, putting coshx and sinhx in exponential form doesnt seem to help...
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    (Original post by MEPS1996)
    Find the integral of 1/(coshx+asinh(x))dx distinguishing between various possibilities of a. Not a clue how to start on this one, putting coshx and sinhx in exponential form doesnt seem to help...
    That looks pretty horrible tbh.

    I would start with the exponential form, multiply top and bottom by e^x, then think about what sort of substiution might help you out.
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    (Original post by davros)
    That looks pretty horrible tbh.

    I would start with the exponential form, multiply top and bottom by e^x, then think about what sort of substiution might help you out.
    ikr, i did this then tried u=e^x, but cant split it into partial fractions because the i cant see how the denominator factorises
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    (Original post by MEPS1996)
    Find the integral of 1/(coshx+asinh(x))dx distinguishing between various possibilities of a. Not a clue how to start on this one, putting coshx and sinhx in exponential form doesnt seem to help...
    For a=1 or a=-1 it is simple
    a=1

    \displaystyle \int \frac{cosh^2 x-sinh^2 x}{cosh x +sinh x} dx=\int \left (cosh x - sinh x\right ) dx

    similarly for a=-1

    \displaystyle \int \left (cosh x + sinh x\right ) dx

    will be the integral.

    for a=0

    \displaystyle \int \frac{1} {cosh x} dx

    substituting t=tanh x
    then
    \displaystyle x=artanh t \rightarrow  dx=\frac{1}{1-t^2} dt
    and
    \displaystyle cosh^2 x=\frac{1}{1- tanh^2 x}=\frac{1}{1-t^2} \rightarrow cosh x=\frac{1}{\sqrt{1-t^2}}

    \displaystyle \int \frac{1}{\frac{1}{\sqrt{1-t^2}}}\cdot \frac{1}{1-t^2} dt=\int \frac{1}{\sqrt{1-t^2}} dt=-i \cdot ar cosh t+C=-i \cdot arcosh \left (tanh x\right ) +C
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    (Original post by MEPS1996)
    Find the integral of 1/(coshx+asinh(x))dx distinguishing between various possibilities of a. Not a clue how to start on this one, putting coshx and sinhx in exponential form doesnt seem to help...
    For the general case, put into exponential form, multiply top and bottom by e^x, then sub e^x=u and you should have a logarithmic integral.
 
 
 
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