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HogoFogo
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
Mar 6, 2009 2:03 PM
Answers · 3
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)
March 7, 2009
OMG!!!!!!!!!!
March 7, 2009
/ 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
March 7, 2009
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HogoFogo
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Czech, English, German, Slovak
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