- #1
DryRun
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Homework Statement
[tex]\int \frac{r^3-r^5}{\sqrt{1-r^2}}\,.dr[/tex]
The attempt at a solution
The presence of [itex]\sqrt{1-r^2}[/itex] suggests that i use the substitution r=sinθ
The integrand becomes: [tex]\frac{\sin^3\theta-\sin^5\theta}{\cos\theta}[/tex]
[tex]\frac{dr}{d\theta}=\cos\theta[/tex]
[tex]\int \frac{r^3-r^5}{\sqrt{1-r^2}}\,.dr=\int \sin^3\theta-\sin^5\theta\,.d\theta=\int\sin^3\theta\,.d\theta-\int\sin^5\theta\,.d\theta[/tex]
Using the substitution u=cosθ
[tex]\int\sin^3\theta\,.d\theta=\frac{\cos^3\theta}{3}-\cos\theta[/tex]
[tex]\int\sin^5\theta\,.d\theta=-\cos \theta+\frac{2\cos^3 \theta}{3}+\frac{\cos^5 \theta}{5}[/tex]
[tex]\int\sin^3\theta\,.d\theta-\int\sin^5\theta\,.d\theta=-\frac{\cos^3 \theta}{3}-\frac{\cos^5 \theta}{5}[/tex]
[tex]\int \frac{r^3-r^5}{\sqrt{1-r^2}}\,.dr[/tex]
The attempt at a solution
The presence of [itex]\sqrt{1-r^2}[/itex] suggests that i use the substitution r=sinθ
The integrand becomes: [tex]\frac{\sin^3\theta-\sin^5\theta}{\cos\theta}[/tex]
[tex]\frac{dr}{d\theta}=\cos\theta[/tex]
[tex]\int \frac{r^3-r^5}{\sqrt{1-r^2}}\,.dr=\int \sin^3\theta-\sin^5\theta\,.d\theta=\int\sin^3\theta\,.d\theta-\int\sin^5\theta\,.d\theta[/tex]
Using the substitution u=cosθ
[tex]\int\sin^3\theta\,.d\theta=\frac{\cos^3\theta}{3}-\cos\theta[/tex]
[tex]\int\sin^5\theta\,.d\theta=-\cos \theta+\frac{2\cos^3 \theta}{3}+\frac{\cos^5 \theta}{5}[/tex]
[tex]\int\sin^3\theta\,.d\theta-\int\sin^5\theta\,.d\theta=-\frac{\cos^3 \theta}{3}-\frac{\cos^5 \theta}{5}[/tex]
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