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After making the correct trig substitution what does the integral of

$\dfrac{1}{\sqrt{9-x^2}} dx$ reduce to without solving the equation?

I reduced it down to the integral of $3\cos(\theta)d\theta$ is this correct?

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Use the substitution $x = 3\sin \theta$. Then $dx = 3\cos \theta$. Your integral will reduce to $$\int \frac {3\cos \theta}{3\cos \theta}\,d\theta = \int 1\cdot \,d\theta=\int d\theta = \theta+C$$

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  • $\begingroup$ there is something wrong with ur coed it only shows dtheta there puu.sh/amxeT/6b99878494.png $\endgroup$ – Panthy Jul 22 '14 at 18:19
  • $\begingroup$ I see there exactly what I see here. (except the "Use the substitution... part) $\endgroup$ – Namaste Jul 22 '14 at 18:20
  • $\begingroup$ $3\cos \theta$ appears in numerator and denominator, and therefore cancel, leaving only the integral of $\,d\theta$. $\endgroup$ – Namaste Jul 22 '14 at 18:21
  • $\begingroup$ ohhhhhhhhhhhhhhhhhhhhhh $\endgroup$ – Panthy Jul 22 '14 at 18:23
  • $\begingroup$ Since you aren't asked to solve, the reduced form of the integral is simply $\int d\theta$. Of course, it integrates to $\theta + C$, and back substituting, we need to solve for $\theta$ in terms of $x$: $$x=3\sin \theta \implies \theta = \sin^{-1}\left(\frac x3\right)$$ $\endgroup$ – Namaste Jul 22 '14 at 18:26
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Performing trig. substitution of $x=3 \sin \theta$ (so $dx=3\cos \theta \, d\theta$) gives $$\int\frac{1}{\sqrt{9-x^2}} \, dx=\int\frac{1}{\sqrt{9-9 \sin^2 \theta}} \, (3 \cos \theta \, d\theta)=\int\frac{1}{|3 \cos \theta|} \, (3 \cos \theta \, d\theta) = \int 1 \, d\theta$$ which is equal to $\theta + C$, or $\displaystyle\sin^{-1} \left( \frac{x}{3} \right) + C$ after back-substitution.

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You need to differentiate $x$ with respect to $\theta$. If you have $x=3\sin\theta$, then you will have $dx=3\cos\theta d\theta.$ Putting it together gives $$\int d\theta$$ (everything cancels)

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