1. trig integral

$\int{sec^4xtan^4x}$

I know the answer is $\frac{1}{9}tan^9x+\frac{1}{7}tan^7x +C$

$\int{{sec^4}3x{tan^4}3x}$

$\frac{1}{9}{tan^9}3x+\frac{1}{7}{tan^7}3x +C$

2. Originally Posted by gammaman
$\int{sec^4xtan^4x}$

I know the answer is $\frac{1}{9}tan^9x+\frac{1}{7}tan^7x +C$

$\int{{sec^4}3x{tan^4}3x}$

$\frac{1}{9}{tan^9}3x+\frac{1}{7}{tan^7}3x +C$
No, it would be 1/3 of that answer plus the constant term, because you are using u-substitution with u = 3x, so du = 3dx inside the integral, hence you must divide by 3 in front of the integral sign. good luck!!

3. Hello, gammaman!

Good thinking . . . but it's slightly wrong.

$\int \sec^4\!x\tan^4\!x\,dx \quad\Rightarrow\quad \tfrac{1}{9}\tan^9\!x+\tfrac{1}{7}\tan^7\!x +C$

But what if I had: . $\int \sec^4\!3x\tan^4\!3x\,dx$

Would the answer just be: . $\tfrac{1}{9}\tan^9\!3x+\tfrac{1}{7}\tan^7\!3x +C$ . . . . no

Because of the $3x$, we'd use: . $u \:=\:3x\quad\Rightarrow\quad du \:=\:3\,dx \quad\Rightarrow\quad dx \:=\:\tfrac{1}{3}\,du$

Substitute: . $\int\sec^4\!u\tan^4\!u\left(\tfrac{1}{3}\,du\right ) \;=\;\tfrac{1}{3}\int\sec^4\!u\tan^4\!u\,du$

We get the same answer ... with $\tfrac{1}{3}$ in front.