Trigonometric Substitution

[Silvanoshei]

Stuck on cancelling on this problem because I suck at sqrts and is it possible to sqrt the terms seperately?

\(\displaystyle ∫\frac{x^{3}}{\sqrt{4-x^{2}}}dx\)

\(\displaystyle ∫\frac{x^{3}}{\sqrt{2^{2}-x^{2}}}dx \rightarrow x = 2sin \theta \rightarrow dx = 2cos \theta d \theta\)

\(\displaystyle ∫\frac{2^{3}\sin^{3}\theta}{\sqrt{4-4\sin^{2}\theta}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle 8∫\frac{\sin^{3}\theta}{\sqrt{4(1-\sin^{2}\theta)}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle 8∫\frac{\sin^{3}\theta}{\sqrt{4\cos^{2}\theta}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle \sqrt{4} \rightarrow \sqrt{\cos^{2}\theta} = 2\cos\theta?\)

 

[Deleted member 4993]

Stuck on cancelling on this problem because I suck at sqrts and is it possible to sqrt the terms seperately?

\(\displaystyle ∫\frac{x^{3}}{\sqrt{4-x^{2}}}dx\)

\(\displaystyle ∫\frac{x^{3}}{\sqrt{2^{2}-x^{2}}}dx \rightarrow x = 2sin \theta \rightarrow dx = 2cos \theta d \theta\)

\(\displaystyle ∫\frac{2^{3}\sin^{3}\theta}{\sqrt{4-4\sin^{2}\theta}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle 8∫\frac{\sin^{3}\theta}{\sqrt{4(1-\sin^{2}\theta)}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle 8∫\frac{\sin^{3}\theta}{\sqrt{4\cos^{2}\theta}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle \sqrt{4} * \sqrt{\cos^{2}\theta} = 2\cos\theta?\)..........................Correct

.

 

[Silvanoshei]

Is this correct?

\(\displaystyle ∫\frac{x^{3}}{\sqrt{4-x^{2}}}dx\)

\(\displaystyle ∫\frac{x^{3}}{\sqrt{2^{2}-x^{2}}}dx \rightarrow x = 2sin \theta \rightarrow dx = 2cos \theta d \theta\)

\(\displaystyle ∫\frac{2^{3}\sin^{3}\theta}{\sqrt{4-4\sin^{2}\theta}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle 8∫\frac{\sin^{3}\theta}{\sqrt{4(1-\sin^{2}\theta)}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle 8∫\frac{\sin^{3}\theta}{\sqrt{4\cos^{2}\theta}} * \frac{2cos \theta}{1} * d \theta\)

\(\displaystyle 8∫\sin^{3}\theta * d \theta\)

\(\displaystyle 8∫\sin^{2}\theta * sin \theta * d \theta\)

\(\displaystyle 8∫(1- \cos^{2}\theta) * sin \theta * d \theta \rightarrow u = cos \theta \rightarrow -du = sin \theta d \theta\)

\(\displaystyle -8∫(1-u^{2})du\)

\(\displaystyle -8(u-\frac{u^{3}}{3}) + C\)

\(\displaystyle -8(cos \theta - \frac{(cos \theta)^{3}}{3}) + C\)

\(\displaystyle x=2sin \theta \rightarrow \frac{x}{2}=sin \theta\)

\(\displaystyle ?^{2} + x^{2} = 2^{2} \rightarrow ? = \sqrt{4-x^{2}} \rightarrow cos \theta = \frac{\sqrt{4-x^{2}}}{2}\)

\(\displaystyle = \rightarrow -8((\frac{\sqrt{4-x^{2}}}{2})-\frac{1}{3}(\frac{\sqrt{4-x^{2}}}{3})^{3}) + C \)

 

[Deleted member 4993]

Looks good to me....

 

[lookagain]

\(\displaystyle ?^{2} + x^{2} = 2^{2} \rightarrow ? = \sqrt{4-x^{2}} \rightarrow cos \theta = \frac{\sqrt{4-x^{2}}}{2}\)

\(\displaystyle = \rightarrow -8((\frac{\sqrt{4-x^{2}}}{2})-\frac{1}{3}(\frac{\sqrt{4-x^{2}}}{3})^{3}) + C \ \ \ \)There is a typo. In this line under the second radical, the denominator is supposed to be a "2," not a "3."

.