# Thread: need to check if this is right

1. ## need to check if this is right

have to integrate this
(3cos4x-2sec^2(3x)+tan x) dx

now i get
3sin4x-2tan3x-3x+ln secx
is this right or do i have to treat each part as like a function of a function
and say like for the first bit
3=U and cos4x=V
and then use du and dv and work it all out then put it back togther
ty

2. Originally Posted by alpharebel10
have to integrate this
(3cos4x-2sec^2(3x)+tan x) dx

now i get
3sin4x-2tan3x-3x+ln secx
is this right or do i have to treat each part as like a function of a function
and say like for the first bit
3=U and cos4x=V
and then use du and dv and work it all out then put it back togther
ty
in first one and the second see this you will find your mistake

$\displaystyle \int sin(6x)dx=\frac{-cos(6x)}{6}+c$

the final one is
$\displaystyle \int \frac{f'(x)}{f(x)} = ln(f(x)) +c$

3. ## thanks for the help

can u just check if this is right i hope it is ty
$\displaystyle int 3cos4x-2sec^2(3x)+tanx$

$\displaystyle frac{3sin4x}{4}-frac{2tan3x}{3}+-lncosx$

4. Originally Posted by alpharebel10
can u just check if this is right i hope it is ty
$\displaystyle int 3cos4x-2sec^2(3x)+tanx$

$\displaystyle \frac{3sin4x}{4}-\frac{2tan3x}{3}+-lncosx$

[tex]\frac{3sin4x}{4}-\frac{2tan3x}{3}+-lncosx[ /math]

[tex]\int 3cos4x-2sec^2(3x)+tanx[ /math]

you solution is correct but you forget to put \ before the frac to present your solution in latex