integral problem: int (sin(3t) 9sec^2(3t))/3 dt

[degreeplus]

I need help solving and understanding this integral

$\displaystyle \int {\frac{{\sin 3t(9\sec ^2 3t)}}{3}}$

the answer is -csc 3t but what I get is -sec 3t

Here is what I did:

I set the $\displaystyle (9\sec^2 3t)$ to $\displaystyle \frac{{9}}{cos^2 3t}$

then I multiplied it by sin 3t to get $\displaystyle 3\int{\frac{{sin 3t}}{cos^2 3t}}$ and then turned it into $\displaystyle \int {3\frac{{\sin 3t}}{{\cos 3t}}} \bullet \frac{1}{{\cos 3t}}$

then this becomes $\displaystyle 3\int{tan 3t \bullet sec 3t}$

and I get -sec 3t

Where did I go wrong?

Thanks for any help in advance

 

[galactus]

Re: integral problem

The solution is indeed $\displaystyle sec(3t)$. You are correct, except for the negative.

 

[Deleted member 4993]

I need help solving and understanding this integral

$\displaystyle \int {\frac{{\sin 3t(9\sec ^2 3t)}}{3}}$

the answer is -csc 3t but what I get is -sec 3t

Here is what I did:

I set the $\displaystyle (9\sec^2 3t)$ to $\displaystyle \frac{{9}}{cos^2 3t}$

then I multiplied it by sin 3t to get $\displaystyle 3\int{\frac{{sin 3t}}{cos^2 3t}}$

you could do substitution here (coming to same answer - except for the sign as galactus pointed out)

u = cos(t)

du = - sin(t) dt

Then integrand turns into -du/u^2

whose antiderivative is (1/u + C = ) sec(t) + C



and then turned it into $\displaystyle \int {3\frac{{\sin 3t}}{{\cos 3t}}} \bullet \frac{1}{{\cos 3t}}$

then this becomes $\displaystyle 3\int{tan 3t \bullet sec 3t}$

and I get -sec 3t

Where did I go wrong?

Thanks for any help in advance