# What is f(x) = int -cos6x -3tanx dx if f(pi)=-1 ?

Sep 8, 2016

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

#### Explanation:

$f \left(x\right) = \int \left(- \cos 6 x - 3 \tan x\right) \mathrm{dx}$

$f \left(x\right) = - \int \cos \left(6 x\right) \mathrm{dx} - 3 \int \tan x \mathrm{dx}$

For the first integral:

$6 x = u$

$\frac{d \left(6 x\right)}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}}$

$6 = \frac{\mathrm{du}}{\mathrm{dx}}$

$\mathrm{dx} = \frac{\mathrm{du}}{6}$

Therefore:

$f \left(x\right) = - \int \cos u \frac{\mathrm{du}}{6} - 3 \int \sin \frac{x}{\cos} x \mathrm{dx}$

$f \left(x\right) = - \frac{1}{6} \int \cos u \mathrm{du} - 3 \int \frac{\left(- \cos x\right) '}{\cos} x \mathrm{dx}$

$f \left(x\right) = - \frac{1}{6} \int \cos u \mathrm{du} + 3 \int \frac{\left(\cos x\right) '}{\cos} x \mathrm{dx}$

$f \left(x\right) = - \frac{1}{6} \sin u + 3 \ln | \cos x | + c$

$f \left(x\right) = - \frac{1}{6} \sin \left(6 x\right) + 3 \ln | \cos x | + c$

Since f(π)=-1

f(π)=-1/6sin(6π)+3ln|cosπ|+c

$- 1 = - \frac{1}{6} \cdot 0 + 3 \ln | - 1 | + c$

$- 1 = 3 \ln 1 + c$

$c = - 1$

Therefore:

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