1. Trigometric Integrals

Evaluate the indefinite integral.

so first i simplified it to ∫(1/4√x²-16)x dx
i took the integral by letting x = 4 secθ, dx= 4secθtanθ
then for √x²-16)= 4tanθ
then i pluged in √x²-16)= 4tanθ and x= 4 secθ for the equation ∫(1/4√x²-16)x dx and my intergral came out to be -1/4cosθ + C but it is wrong i dont get what i have to do next?

2. Originally Posted by emurphy
Evaluate the indefinite integral.

so first i simplified it to ∫(1/4√x²-16)x dx
i took the integral by letting x = 4 secθ, dx= 4secθtanθ
then for √x²-16)= 4tanθ
then i pluged in √x²-16)= 4tanθ and x= 4 secθ for the equation ∫(1/4√x²-16)x dx and my intergral came out to be -1/4cosθ + C but it is wrong i dont get what i have to do next?

$\int\frac{\sqrt{16x^2-256}}{x}\,dx=4\int\frac{\sqrt{x^2-16}}{x}\,dx$

You made a correct choice for $x$: $x=4\sec t, dx=4\sec t\tan t\,dt$

$4\int\frac{4\tan t}{4\sec t}\cdot4\sec t\tan t\,dt=16\int\tan^2t\,dt=16\int(\sec^2t-1)\,dt=16\tan t-16t$

Now sub back in for $t$.

3. Originally Posted by emurphy
Evaluate the indefinite integral.

so first i simplified it to ∫(1/4√x²-16)x dx
i took the integral by letting x = 4 secθ, dx= 4secθtanθ
then for √x²-16)= 4tanθ
then i pluged in √x²-16)= 4tanθ and x= 4 secθ for the equation ∫(1/4√x²-16)x dx and my intergral came out to be -1/4cosθ + C but it is wrong i dont get what i have to do next?

${4}\int \frac{\sqrt{x^2-16}}{x} \ dx = 4\int \sqrt{1-16/x^2} \ dx$

Set $4/x=\sin \theta$, $dx = -4\cot \theta \csc \theta\ d\theta$

$4\int \sqrt{1-\sin^2 \theta} (-4\cot \theta \csc \theta\ d\theta) = -16\int \cos \theta \cot \theta \csc \theta\ d\theta$

$=-16 \int \cos \theta \frac{\cos \theta}{\sin \theta}\frac{1}{\sin \theta}\ d\theta = -16 \int \cot^2 \theta d\theta$

$=16 (\theta+\cot \theta)+ C = 16(\arcsin(4/x)+\cot(\arcsin(4/x)))+C$

Edit : above is better!

4. I dont understand how to get t

5. Originally Posted by emurphy
I dont understand how to get t
If $x=4\sec t$, then $t=\sec^{-1}\left(\frac{x}{4}\right)$