## .Sam. Group Title @tanvidais13 one year ago one year ago

1. .Sam. Group Title

Your question $\int\limits \frac{7}{\sqrt{5+4x-x^2}}$ Try completing the square @tanvidais13

2. tanvidais13 Group Title

wait i don't need to, its factorizable: $7/\sqrt{(1+x)(5-x)}$

3. .Sam. Group Title

no you can't do that because we can't use partial fractions here

4. tanvidais13 Group Title

so we need to complete the square, or do a substitution?

5. .Sam. Group Title

We can try using trig substitution, it might work, so if we complete the square then we got $\int\limits\limits \frac{7}{\sqrt{9-(x-2)^2}} \, dx$ Put it like this, $\int\limits\limits\limits \frac{7}{\sqrt{3^2-(x-2)^2}} \, dx$ Using trig sub, $x=asin(\theta), \text{while} ~~~~~ a^2-x^2 ~~~~~ \text{is} ~~~~~ 3^2-(x-2)^2, \\ \\So \\ \\------------\\\\ x-2=3\sin(\theta) ~~~~~ \theta=\sin^{-1}(\frac{x}{3}) \\ \\ dx=3\cos (\theta) d \theta \\\\-----------\\\\ 7 \int\limits \frac{3\cos \theta}{\sqrt{3^2-(3\sin \theta)^2}}d \theta$

6. .Sam. Group Title

$7 \int\limits \frac{3\cos \theta}{\sqrt{3^2(1-\sin^2 \theta)}}d \theta \\ \\ \text{We know that} \sqrt{ab}=\sqrt{a} \sqrt{b}\\ \\ 7 \int\limits \frac{\cos \theta}{\sqrt{(1-\sin^2 \theta)}}d \theta \\ \\ 7 \int\limits \frac{\cos \theta}{\sqrt{\cos^2 \theta}}d \theta \\ \\7 \int\limits 1 d \theta$

7. .Sam. Group Title

$7 \theta +c \\ \\ 7\sin^{-1}(\frac{x-2}{3}) +c$

8. .Sam. Group Title

Note that theta is$\theta=\sin^{-1}(\frac{u}{3})$ and u=x-2

9. .Sam. Group Title

@tanvidais13 Both completing the square and substitution needed

10. tanvidais13 Group Title

I love you, you are a life saver. Really.

11. .Sam. Group Title

Welcome :)