2

Problem is as follows

Find the anti-derivative of the function $$f(x) = \frac{8}{\sqrt{36-4x^2}}$$

Started Calc 2 last week and we are doing $u$-substitution review. However, no matter what I try, I have no clue how to even begin this problem. $u$-subbing the denominator gets me stuck.

Edit: Thanks for the help everyone! Im definitely going to practice my u-sub more now.

Luke D
  • 159

4 Answers4

9

Hint:

Let $$ x=3\sin(u),\ dx= 3\cos(u)$$

Xander Henderson
  • 32,453
Zach
  • 350
  • It might be a good idea to familiarize yourself with MathJax so that you can properly format your mathematics. Also with respect to this answer (which you created then deleted), it is worth noting that there is a Sandbox for playing with formatting. – Xander Henderson Aug 26 '18 at 16:50
  • Thanks so much!! – Zach Aug 26 '18 at 16:51
  • Good answer Zach. It may be a good idea to explain how you got that substitution, as to me it isn't completely obvious. – Rhys Hughes Aug 26 '18 at 16:57
  • Thanks for the quick comment! But I'm confused from where the 3 comes from. Could you explain the origin/use of it? – Luke D Aug 26 '18 at 16:57
  • I want to be able to factor the 36 out of the square root. Since there is a 4 already present I want to have 4 times some number squared giving me 36. This number is 3 because 3 squared is 9, times 4 is 36! Now we can factor out 36 and simplify! – Zach Aug 26 '18 at 16:59
  • The expression for $dx$ needs a $du$. – Barry Cipra Aug 26 '18 at 17:06
  • Thanks for that hint, it got me pretty far through the problem. Im currently stuck at 4cosu/sqrt(cos^2u) *du – Luke D Aug 26 '18 at 17:06
1

$$\int \frac{8}{\sqrt{36-4x^2}}dx$$

put $x=3 \sin t$ thus $dx=3 \cos t dt $ and $t=\sin^{-1}(x/3)$ $$\int \frac{8\cdot3 \cos t dt }{6\cdot \cos t}=\int 4 dt$$

Deepesh Meena
  • 4,496
1

Factor out $36$ in the radical, and set $u=\dfrac x3$, to obtain a well-known anti-derivative: $$\int\frac 8{\sqrt{36-4x^2}}\,\mathrm dx=\int\frac 4{3\sqrt{1-\bigl(\frac x3\bigr)^2}}\,\mathrm dx=4\int\frac {\mathrm d u}{\sqrt{1-u^2}}=4\arcsin u=\dots$$

Bernard
  • 179,256
0

Hint (with motivational idea): If you know that $$ \arcsin x = \int \frac{1}{\sqrt{1-x^2}} \; \mathrm d x,$$ then you can rewrite your integrand as $$\frac{8}{\sqrt{36-4x^2}} = \frac{2}{\sqrt{9-x^2}} = \frac{2}{ \sqrt{9} \sqrt{1-\left(\frac{x}{3} \right)^2}}.$$ Finally, a linear substitution finishes the integral.