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

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Answer 1 · 2018-04-21T23:39:49

$\int (x e^{x} - sec 7 x tan 7 x) d x = x e^{x} - e^{x} - \frac{1}{7} sec (7 x) + C$ $int(xe^x-sec7xtan7x)dx=xe^x-e^x-1/7sec(7x)+C$

So, we want

$\int (x e^{x} - sec 7 x tan 7 x) d x$ $int(xe^x-sec7xtan7x)dx$ . We can split up across the difference, yielding the following two integrals:

$\int x e^{x} d x - \int sec 7 x tan 7 x d x$ $intxe^xdx-intsec7xtan7xdx$

For $\int x e^{x} d x$ $intxe^xdx$ , we will use Integration by Parts, making the following selections:

$u = x$ $u=x$
$d u = d x$ $du=dx$
$d v = e^{x} d x$ $dv=e^xdx$
$v = \int e^{x} d x = e^{x}$ $v=inte^xdx=e^x$

$u v - \int v d u = x e^{x} - \int e^{x} d x$ $uv-intvdu=xe^x-inte^xdx$

$= x e^{x} - e^{x}$ $=xe^x-e^x$

For $\int sec 7 x tan 7 x d x$ $intsec7xtan7xdx$ , let's first make a simple substitution to clean things up:

$u = 7 x$ $u=7x$
$d u = 7 d x$ $du=7dx$
$\frac{1}{7} d u = d x$ $1/7du=dx$

Then, we have the common integral

$\frac{1}{7} \int sec u tan u d u = \frac{1}{7} sec u = \frac{1}{7} sec (7 x)$ $1/7intsecutanudu=1/7secu=1/7sec(7x)$

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

$\int (x e^{x} - sec 7 x tan 7 x) d x = x e^{x} - e^{x} - \frac{1}{7} sec (7 x) + C$ $int(xe^x-sec7xtan7x)dx=xe^x-e^x-1/7sec(7x)+C$

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