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please help me with "easy" integrals

hi everybody,
can somebody help me with these integrals? i would be really thankful for any advice.

e^(arcsin x) dx

e^(arccos x) dx

e^(arctg x) dx

e^(arccotg x) dx

For learning: Chinese (Mandarin)
Base language: English
Category: Uncategorized

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    / e^(arcsinx)dx = x * e^(arcsinx) - / x * e^(arcsinx)/sqrt(1-x^2) dx
    = x * e^(arcsinx) + / 1/2 * e^(arcsinx) * 2 d sqrt(1 - x^2)
    = x * e^(arcsinx) + e^(arcsinx) * sqrt(1 - x^2) - / sqrt(1 - x^2)/ sqrt(1 - x^2) e^(arcsinx) dx
    = x * e^(arcsinx) + e^(arcsinx) * sqrt(1 - x^2) - / e^(arcsinx)dx
    then can get: / 2e^(arcsinx)dx = x * e^(arcsinx) + e^(arcsinx) * sqrt(1 - x^2) + C'
    then: / e^(arcsinx)dx = 1/2 * e^(arcsinx) [x + sqrt(1 - x^2)] + C

    or : let arcsinx = u, (-pi/2 < u < pi/2 ), then x = sinu, dx = cosu du ,
    / e^(arcsinx) dx = / e^u * cos u * du
    = / cos u * d e^u
    = cos u * e^u + / e^u sin u du
    = cos u * e^u + / sin u d e^u
    = cos u * e^u + sin u * e^u - / e^u cos u du
    then can get : 2 / e^u cos u du = cos u * e^u + sin u * e^u + C'
    then : / e^u cos u du =1/2 [ cos u + sin u * ] * e^u + C
    = 1/2 [ sqrt(1 - x^2) + x] * e^(arcsinx) + C

    OMG!!!!!!!!!!
    the derivative: 1/2 *( [sqrt(1-x^2)+x] * e^(arcsinx) )'
    = 1/2 *[ 1/2 *(-2x) /sqrt(1 - x^2)+1 + x/sqrt(1 - x^2) +1 ] * e^(arcsinx)
    = 1/2 *2e^(arcsinx)
    = e^(arcsinx)

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