How do you find the integral of $xe^x - sec(7x)tan(7x) dx$ ?

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Answer 1 · 2018-04-21T23:39:49

$int(xe^x-sec7xtan7x)dx=xe^x-e^x-1/7sec(7x)+C$

So, we want

$int(xe^x-sec7xtan7x)dx$ . We can split up across the difference, yielding the following two integrals:

$intxe^xdx-intsec7xtan7xdx$

For $intxe^xdx$ , we will use Integration by Parts, making the following selections:

$u=x$
$du=dx$
$dv=e^xdx$
$v=inte^xdx=e^x$

$uv-intvdu=xe^x-inte^xdx$

$=xe^x-e^x$

For $intsec7xtan7xdx$ , let's first make a simple substitution to clean things up:

$u=7x$
$du=7dx$
$1/7du=dx$

Then, we have the common integral

$1/7intsecutanudu=1/7secu=1/7sec(7x)$

Combining our integrals together and putting in the constant of integration, we get

$int(xe^x-sec7xtan7x)dx=xe^x-e^x-1/7sec(7x)+C$

How do you find the integral of xe^x - sec(7x)tan(7x) dx xe^x - sec(7x)tan(7x) dx?