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anonymous 4 years ago Use a trigonometric substitution involving tangent to write √(961 + x^2) as a function of theta in the first quadrant without a radical So confused pleaseeee help

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1. anonymous
2. anonymous

would it help to know that $$\sqrt{961}=31$$?

3. anonymous

put $$x=32\tan(\theta)$$ so $$x^2=961\tan^2(\theta)$$ and you get $\sqrt{961+x^2}=\sqrt{961+961\tan^2(\theta)}=\sqrt{961(1+\tan^2(\theta)}=31\sqrt{1+\tan^2(\theta)}$

4. anonymous

sorry i meant $x=31\tan(\theta)$

5. anonymous

and now a miracle occurs, namely that $1+\tan^2(\theta)=\sec^2(\theta)$ so you can get rid of the radical

6. anonymous

when you see $\sqrt{a^2+x^2}$ use $x=a\tan(\theta)$ and when you see $\sqrt{a^2-x^2}$ use $x=a\sin(\theta)$

7. anonymous

we can do another if you like

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