What is $f (x) = \int - cos 6 x - 3 tan x d x$ $f(x) = int -cos6x -3tanx dx$ if $f (π) = - 1$ $f(pi)=-1$ ?

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Answer 1 · 2016-09-08T08:15:23

Answer is:

$f (x) = - \frac{1}{6} sin (6 x) + 3 ln | cos x | - 1$ $f(x)=-1/6sin(6x)+3ln|cosx|-1$

$f (x) = \int (- cos 6 x - 3 tan x) d x$ $f(x)=int(-cos6x-3tanx)dx$

$f (x) = - \int cos (6 x) d x - 3 \int tan x d x$ $f(x)=-intcos(6x)dx-3inttanxdx$

For the first integral:

$6 x = u$ $6x=u$

$\frac{d (6 x)}{d x} = \frac{d u}{d x}$ $(d(6x))/(dx)=(du)/dx$

$6 = \frac{d u}{d x}$ $6=(du)/dx$

$d x = \frac{d u}{6}$ $dx=(du)/6$

Therefore:

$f (x) = - \int cos u \frac{d u}{6} - 3 \int \frac{sin x}{cos x} d x$ $f(x)=-intcosu(du)/6-3intsinx/cosxdx$

$f(x)=-1/6intcosudu-3int((-cosx)')/cosxdx$

$f(x)=-1/6intcosudu+3int((cosx)')/cosxdx$

$f(x)=-1/6sinu+3ln|cosx|+c$

$f(x)=-1/6sin(6x)+3ln|cosx|+c$

Since $f(π)=-1$

$f(π)=-1/6sin(6π)+3ln|cosπ|+c$

$-1=-1/6*0+3ln|-1|+c$

$-1=3ln1+c$

$c=-1$

Therefore:

$f(x)=-1/6sin(6x)+3ln|cosx|-1$

What is f(x) = int -cos6x -3tanx dxf(x)=∫−cos6x−3tanxdxf(x) = int -cos6x -3tanx dx if f(pi)=-1 f(π)=−1f(pi)=-1 ?