1. ## integration check!

$\displaystyle \int (4sec^2(5x-6)+7sec(9-8x)tan(9-8x)) dx$

$\displaystyle = \frac {-7}{8}sec(9 - 8x) - \frac {4}{5}tan(6 - 5x)$

is this correct? thanks

2. Originally Posted by jvignacio
$\displaystyle \int (4sec^2(5x-6)+7sec(9-8x)tan(9-8x)) dx$

$\displaystyle = \frac {-7}{8}sec(9 - 8x) - \frac {4}{5}tan(6 - 5x)$

is this correct? thanks
you are forgetting a +C, but other than that, it looks good.

i will say that you have something weird here though. it seems counter-intuitive that you would integrate backwards here. and that you would change the angle of the tangent by a factor of -1 and take the negative of the function

had i done the problem, it would look like $\displaystyle \frac 45 \tan (5x - 6) - \frac 78 \sec (9 - 8x) + C$

3. Originally Posted by Jhevon
you are forgetting a +C, but other than that, it looks good.

i will say that you have something weird here though. it seems counter-intuitive that you would integrate backwards here. and that you would change the angle of the tangent by a factor of -1 and take the negative of the function

had i done the problem, it would look like $\displaystyle \frac 45 \tan (5x - 6) - \frac 78 \sec (9 - 8x) + C$
very true. thank you