How do you integrate #int e^x tan sqrtx dx # using integration by parts?

1 Answer
Jorge G. Pires
May 14, 2016

You have to apply the definition: #intudv=uv - intvdu #. Nonetheless, this integral does seem solvable analytically.

Explanation:

Integration by parts

If you apply it once:

Choose: #dv=exp(x)dx#, and #u=tansqrt(x)#
Which gives you: #v=exp(x)#, and #du=sec^2sqrt(x)/(2sqrt(x))dx#

Thus:

#intexp(x)tansqrt(x)dx= exp(x)*sec^2sqrt(x)/(2sqrt(x)) - intexp(x)sec^2sqrt(x)/(2sqrt(x))dx#

Unfortunately, the more you go, the more complicate it gets, neither if you make #u=exp(x)#, and #dv=tansqrt(x)dx#, the exponential "locks" the integrals, making them cumbersome. Maybe this integral is no solvable by integration by part, I am not sure even analytically.

Learn more: https://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions, https://en.wikipedia.org/wiki/Lists_of_integrals, https://en.wikipedia.org/wiki/Integration_by_parts

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A possible way to solve it is by Taylor expansion, expand #tansqrt(x)#, which is the problematic part of the integral. I will try to add a second answer using this method.

Taylor expansion to tangent

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