Question

Another calculus problem.. >_<

Use the substitution x=8tan() to evaluate the indefinite integral:

72/(x^2 sqrt(x^2+64))

Thanks! :D

Answer 1

waht is 8tan(), do you ean for theta?

Answer 2

*mean

Answer 3

Yeah...

Answer 4

hint* if x = 8tan()
then x^2 = 64tan^2()

and remember tan^2() + 1 = Sec^2()

Answer 5

so first try plugging 64tan^2() for all x^2

Answer 6

Yeah.. That's what I did.. But I got stuck after that.. When I have \[(9/64)\int\limits 1/(\tan^2\theta \sec \theta)\]

Answer 7

remember your indefinite integral needs a (dx)

if x = 8tan()
then dx = 8sec^2()d()

Answer 8

\[\int\limits_{}^{}(72/x^2\sqrt{x^2 + 64} )dx = 9/8\int\limits_{}^{}[/\tan^2\theta \sec \theta]\sec^2\theta\]

Answer 9

\[ 9/8\int\limits_{}^{}[1/(\tan^2\theta)(\sec \theta)] \sec^2\theta d \theta = 9/8\int\limits_{}^{}(\sec \theta / \tan^2\theta) d \theta\]

\[= 9/8\int\limits_{}^{}\csc \theta \cot \theta d \theta\]

\[= -9/8 \csc \theta\]

Answer 10

Ohh.. That makes sense! Thank you so much! :D

Answer 11

haha i remember one of my friends said " you don't need trig for calculus". He didnt pass calc 2

Answer 12

Ohh wow.. That's really funny.. >_<