Question

What is the integral of \[\sqrt{9-{{x}^{2}}}\]?

Answer 1

Hint: In this problem, we have to find the integral of the given square root. Here we can use the substitution method and substitute the value of x as \[3\sin t\], we can then simplify the steps inside the integral using trigonometric formulae. We can then integrate the problem, and replace the t value as x.

Complete step by step answer:
We know that the given integral is,
 \[\int{\sqrt{9-{{x}^{2}}}}dx\]
We can now use the substitution method to integrate the given problem.
We can substitute for the x values,
Let \[x=3\sin t\],
We can now differentiate the above substitution, we get
\[\Rightarrow dx=3\cos tdt\]
We can now replace the x term with t, from the above substitutions, we get
 \[= \int{3\sqrt{1-{{\sin }^{2}}t}}\times 3\cos tdt\]
We can now simplify the above terms, by using the trigonometric formula \[\sqrt{1-{{\sin }^{2}}t}=\cos t\] and taking the constant term outside, we get
\[\Rightarrow 9\int{\cos t\times \cos tdt=9\int{{{\cos }^{2}}tdt}}\]
We can now write the above step using the formula \[\cos 2A=2{{\cos }^{2}}A-1\], we get,
\[= \dfrac{9}{2}\int{\left( 1+\cos 2t \right)dt}\]
We can now integrate the above step, we get
\[= \dfrac{9}{2}\left( t+\dfrac{1}{2}\sin 2t \right)+C\]
We can now write the above step in terms of x, we get
\[= \dfrac{9}{2}\left( {{\sin }^{-1}}\left( \dfrac{x}{3} \right)+\dfrac{x}{3}\sqrt{1-{{\left( \dfrac{x}{3} \right)}^{2}}} \right)\]

Therefore, the value of the given integral is, \[\int{\sqrt{9-{{x}^{2}}}}dx=\dfrac{9}{2}\left( {{\sin }^{-1}}\left( \dfrac{x}{3} \right)+\dfrac{x}{3}\sqrt{1-{{\left( \dfrac{x}{3} \right)}^{2}}} \right)\].

Note: Students make mistakes while substitution, where we have to substitute the correct terms to get the final answer correct. We should always remember that the trigonometric formula used in this problem are \[\cos 2A=2{{\cos }^{2}}A-1\] and \[\sqrt{1-{{\sin }^{2}}t}=\cos t\]. We should also remember that we have to replace the t terms to x terms at the final step.