I am confused about how $\sqrt{\sin^2 6t +\cos^2 6t + (6\ln(\cos t))^2}$ simplifies to $36 \sec t$

Question

$$\int (\sin^2 6t +\cos^2 6t + (6\ln(\cos t))^2)^{1/2} dt $$ $$\int (36 (\sin^2 6t + \cos^2 6t + \tan^2 t) )^{1/2} dt $$ $$\int 36 \sec t\,dt $$

the part that I do not get is when $6\ln(\cos t))^2$ is turned to $\tan^2 t$ with $36$ factored out. I'd appreciate any help. thank you so much!

Answer 1

It is incorrect to go from $(6\log(\cos t))^2$ to something with no $\log$.

Answer 2

I'm going to make a guess here based on the "arc length" tag. I think you have tried to compute an arc length using $$\int (x^2+y^2+z^2)^{1/2}\,dt$$ when it should be $$\int \Bigl(\Bigl(\frac{dx}{dt}\Bigr)^2+\Bigl(\frac{dy}{dt}\Bigr)^2+\Bigl(\frac{dz}{dt}\Bigr)^2\Bigr)^{1/2}\,dt\ .$$ This fixes things up, mostly: there is still one minor error which I'll leave you to find.