Integration: Find length of curve using NINT
Here are the questions -
For question 4, part (b) gives a unit circle. But I'm unable to proceed with parts (a) and (c), since the curve is double valued for -0.5
Also, for question 6, integration of x cos x gives me problems. Could someone help me by guiding me through the steps for these 2?
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$\begingroup$Question 4:
Integral: $$\int_{-\frac{1}{2}}^{\frac{1}{2}}\sqrt{1-y^2}dy$$
To solve, use substitution $y=\sin(u)$
We have $\frac{dy}{du}=\cos(u)$, so $dy=\cos(u)du$
We also replace the limits giving $u_2=\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}$ and $u_1=\sin^{-1}(-\frac{1}{2})=-\frac{\pi}{6}$
Substituting in $y=\sin(u)$, we get
$$\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\sqrt{1-\sin^2(u)}\cos(u)du=\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}cos^2(u)du$$
Then use the identity $2\cos^2(u)-1=\cos(2u)$ or $\cos^2(u)=\frac{1}{2}(cos(2u)-1)$
This gives the integral $$\frac{1}{2}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}(\cos(2u)+1)du$$
Which is $$\left.\frac{1}{2}(u+\frac{1}{2}\sin(2u))\right|_{-\frac{\pi}{6}}^{\frac{\pi}{6}}$$
Substituting in our values for $u$, we get $$\frac{\pi}{6}+\frac{\sqrt{3}}{4}$$
Question 6:
Integral: $$\int_0^\pi \sin(x)-x\cos(x)dx$$
Hint: After you seperate the integral into $$\int_0^\pi \sin(x)-\int_0^\pi x\cos(x)dx$$
For the $$\int_0^\pi x\cos(x)dx$$ part, use integration by parts with one part $x$ and the other part $\cos(x)$
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