What is $f (x) = \int - cos 6 x - 3 tan x + cot (\frac{x}{3}) d x$ $f(x) = int -cos6x -3tanx+cot(x/3) dx$ if $f (π) = - 2$ $f(pi)=-2$ ?

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Answer 1 · 2016-05-27T04:47:10

$f (x) = \int (- cos 6 x - 3 tan x + cot (\frac{x}{3})) d x$ $f(x) = int( -cos6x -3tanx+cot(x/3)) dx$

$\Rightarrow f (x) = - \int cos 6 x d x - 3 \int tan x d x + \int cot (\frac{x}{3}) d x$ $=> f(x) = -intcos6xdx -3inttanxdx+intcot(x/3) dx$

$=> f(x) = -1/6sin6x -3ln|secx|+3ln|sin(x/3)| +c ...color(red)"(1)"$

,where c= integration constant

Now imposing the given condition $f(pi)=-2$

$-1/6sin(6xxpi) -3ln|secpi|+3ln|sin(pi/3)| +c=-2$

$1/6xx0-3xx0+3xxln(sqrt3/2)+c=-2$

$c=-2-3ln(sqrt3/2)$

Inserting the value of c in $equation color(red)"(1)"$ we have

$f(x) = -1/6sin6x -3ln|secx|+3ln|sin(x/3)| -2-3ln(sqrt3/2)$

What is f(x) = int -cos6x -3tanx+cot(x/3) dxf(x)=∫−cos6x−3tanx+cot(x3)dxf(x) = int -cos6x -3tanx+cot(x/3) dx if f(pi)=-2 f(π)=−2f(pi)=-2 ?