I'm trying to compute the length of a curve.
I'm left with the definite integral $$ \int \sqrt{1^2 + 4t^2 + 9t^4 }dt $$ which i reduced to $$ \int 2t\sqrt{1+(3 t/2)^2}dt $$
I have a feeling I'm supposed to replace 3/2 t with -sint but I'm not completely sure. if i made it negative it would just square the negative...
but I'm stuck here.. I'm probably supposed to use a rule from calc 2 i forgot like u substitution but I can't remember how.. I've been looking up examples but I can't seem to apply it to here..
So how do I proceed to take this integral?
Assuming you have manipulated the first equation correctly,
$$\int 2t \sqrt{1+(\frac{3t}{2})^2}dt$$
$$t=\frac{2}{3}\tan x$$ $$dt=\frac{2}{3}\sec ^2xdx$$ $$\int \frac{4}{3}\tan x \sqrt{1+\tan^2 x} *\frac{2}{3}\sec ^2x dx$$ $$\frac{8}{9}\int \tan x \sec^3 x dx$$ $$\frac{8}{27}(\sec^3 x) +C$$
Then substitute $t$ back into the equation by writing $x$ in terms of $t$
$$\frac{9}{4} t^2=\tan^2x=\sec^2x-1$$ $$\sec^2x=\frac{9}{4}t^2+1$$
So the solution of the integral is
$$\frac{8}{27}(\frac{9}{4}t^2+1)^{\frac{3}{2}} +C$$
However, as pointed out by user222031, you have manipulated the first equation incorrectly.