thx

2. \displaystyle \begin{aligned} {\displaystyle{\int}}\tan^3\,3x\,\sec^4\,3x\,dx &= {\displaystyle{\int}}\tan^3\,3x\,\sec^2\,3x\,\sec^ 2\,3x\,dx \\ &= {\displaystyle{\int}}\tan^3\,3x\,(1 + \tan^2\,3x)\sec^2\,3x\,dx \\ &= {\displaystyle{\int}}(\tan^3\,3x\, + \tan^5\,3x\,)\sec^2\,3x\,dx \\ &= {\displaystyle{\int}}\tan^3\,3x\,\sec^2\,3x\,dx + {\displaystyle{\int}}\tan^5\,3x\,\sec^2\,3x\,dx \\ \end{aligned}

\displaystyle \begin{aligned} u &= \tan\,3x \\ du &= 3\sec^2\,3x\,dx \\ \frac{1}{3}du &= \sec^2\,3x\,dx \end{aligned}

Can you take it from here?

3. Hello, kenziefc!

We have to guess what the integral is supposed to be.

Here's my approach . . .

$\displaystyle \int\tan^3\!3x\!\cdot\!\sec^4\!3x\!\cdot\!dx$ . . . . . my guess

We have: .$\displaystyle \int \sec^3\!3x\!\cdot\!\tan^2\!3x\!\cdot\!(\sec3x\!\cd ot\!\tan3x\!\cdot\!dx)$

. . . . . $\displaystyle =\;\int\sec^3\!3x\!\cdot\!(\sec^2\!3x - 1)\!\cdot\!(\sec3x\!\cdot\!\tan3x\!\cdot\!dx)$

. . . . . $\displaystyle =\;\int(\sec^5\!3x - \sec^3\!3x)\!\cdot\!(\sec3x \tan3x\,dx)$

Let: $\displaystyle u = \sec3x$

. . $\displaystyle du \,=\,3\sec3x\tan3x\,dx \quad\Rightarrow\quad \sec3x\tan3x\,dx \,=\,\frac{1}{3}\,du$

Substitute: .$\displaystyle \int (u^5-u^3)\cdot\frac{1}{3}\,du \;=\;\frac{1}{3}\int(u^5 - u^3)\,du$

I assume you can finish it . . .

4. thx!!
helped a lot!

5. Another way to do this (without so much guessing): convert to sine and cosine:
$\displaystyle \int tan^3(3x)sec^4(3x)dx= \int \frac{sin^3(3x)}{cos^3(3x)}\frac{1}{cos^4(3x)}dx= \int \frac{sin^3(3x)}{cos^7(3x)}dx$

Now, since that has sin(3x) to an odd power, the standard method is to factor one out for the differential and convert to cos(3x): $\displaystyle \int\frac{sin^3(3x)}{cos^7(3x)}dx= \int\frac{1- cos^2(3x)}{cos^7(3x)}(sin(x)dx)$.

Let u= cos(x) so that du= -sin(x)dx and the integral becomes $\displaystyle -\int\frac{1- u^2}{u^7}du= \int (u^{-5}- u^{-7})du$.

6. We could also do with 'straightforward' substitution by letting $\displaystyle \displaystyle u = \sec^4{3x}$,

Then $\displaystyle \displaystyle \frac{du}{dx} = 12\tan{3x}\sec^4{3x} \Rightarrow dx = \frac{du}{12\tan{3x}\sec^4{3x}}$

Thus $\displaystyle \displaystyle \int\tan^3{3x}\sec^4{3x}\;{dx} = \int\dfrac{\tan^3{3x}\sec^4{3x}}{12\tan{3x}\sec^4{ 3x}}\;{du} = \frac{1}{12}\int\tan^2{3x}\;{du}$.

Now, since $\displaystyle \displaystyle u = \sec^4{3x} = (\tan^2{3x}+1)^2$, we have $\displaystyle \displaystyle \tan^2{3x} = \sqrt{u}-1$.

Thus $\displaystyle \displaystyle \frac{1}{12}\int\tan^2{3x}\;{du} = \frac{1}{12}\int \sqrt{u}-1 \;{du} = ...$

Everyone so generously left something for the OP to do, and so would I.