How do you integrate $int x/ cos^2(x)dx$ ?

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Answer 1 · 2016-09-10T20:28:17

$xtan(x)+ln(abscos(x))+C$

We have:

$I=x/cos^2(x)dx=intxsec^2(x)dx$

We will use integration by parts, which takes the form $intudv=uv-intvdu$ . For the given integral $intxsec^2(x)dx$ , let:

${(u=x,(du)/dx=1,du=dx),(dv=sec^2(x)dx,intdv=intsec^2(x)dx,v=tan(x)):}$

Thus:

$I=xtan(x)-inttan(x)dx$

$I=xtan(x)+int(-sin(x))/cos(x)dx$

Letting $y=cos(x)$ such that $dy=-sin(x)dx$ :

$I=xtan(x)+int(dy)/y$

$I=xtan(x)+ln(absy)$

$I=xtan(x)+ln(abscos(x))+C$

How do you integrate int x/ cos^2(x)dxint x/ cos^2(x)dx?