Peform a trigonometric substitution on the following integral. You do NOT need to do the new (d theta) integral. ((ln(x sqrt(49-x^2)) 5))/(7-xsqrt(49-x^2))

Question

lgbasallote · Answer

trig sub without theta? i don't get it

anonymous · Answer

actually just ignore that part...but thats what it says on my question

anonymous · Answer

is that "5" an exponent?

anonymous · Answer

+5

parthkohli · Answer

If we have $a^2 - x^2$ inside the square root, then substitute $x = a\sin\theta$.

parthkohli · Answer

If we have something in that form, I mean...^

parthkohli · Answer

I find everything confusing without $\LaTeX$.

anonymous · Answer

??

anonymous · Answer

is it this?  (god, please say no!):
$\huge \int \frac{ln(x\sqrt{49-x^2}+5)}{7-x\sqrt{49-x^2}}dx $

lgbasallote · Answer

$$\sqrt{49 - x^2}$$
so your triangle looks like |dw:1347148466479:dw|
so 7\cos \theta = \sqrt{49 - x^2}\]
and $$7 \sin \theta = x$$