91Ó°ÊÓ

First, let's find the derivative of \({\bf r}(t)=\langle 1+\sin t, \cos^2 t\rangle\) with respect to \(t\): \({\bf r}'(t) = \frac{d}{dt} \langle 1+\sin t, \cos^2 t \rangle = \langle \cos t, -2\cos t \sin t \rangle\) Now, we can proceed to the next step.

Next, we need to substitute the parameterized curve \({\bf r}(t)\) and its derivative \({\bf r}'(t)\) into the line integral: \(\int_{C} x^{3} d x+y^{3} d y = \int_{0}^{\frac{\pi}{2}} [(1+\sin t)^{3}(\cos t) + (\cos^2 t)^{3}(-2\cos t \sin t)] dt\) Now, the expression inside the integral is ready to be integrated.

Now, we integrate the expression inside the integral with respect to \(t\), over the interval \([0, \frac{\pi}{2}]\): \(\int_{0}^{\frac{\pi}{2}} [(1+\sin t)^{3}(\cos t) - 2(\cos^4 t)(\sin t \cos t)] dt\) Now, we can use substitution with \(u = 1+\sin t\) and \(v=\cos^2t\). For \(u\), we get \(du=\cos t dt\), so the first part becomes: \(\int u^3du=\int (1+\sin t)^3\cos t dt = \frac{1}{4}u^4=\frac{1}{4}(1+\sin t)^4+C_1\) For \(v\), we get \(dv=-2\sin t \cos t dt\), so the second part becomes: \(\int -2v^2dv=-2\int \cos^4 t \sin t \cos t dt = -\frac{2}{3} v^3 = -\frac{2}{3} \cos^6 t +C_2\) Thus, the overall line integral is: \(\int_{0}^{\frac{\pi}{2}} x^3 dx+y^3dy=\frac{1}{4}(1+\sin t)^4-\frac{2}{3} \cos^6 t +C\) Now, we need to find the definite integral over the interval \([0, \frac{\pi}{2}]\): \(=\left[\frac{1}{4}(1+\sin t)^4\right]_{0}^{\frac{\pi}{2}}-\left[\frac{2}{3} \cos^6 t\right]_{0}^{\frac{\pi}{2}}\) \(= \frac{1}{4}(1+\sin(\frac{\pi}{2}))^4-\frac{2}{3} \cos^6 (\frac{\pi}{2})-\left(\frac{1}{4}(1+\sin(0))^4-\frac{2}{3} \cos^6 (0)\right)\) \(= \frac{1}{4}(2)^4-\frac{2}{3}(0)-\left(\frac{1}{4}(1)^4-\frac{2}{3}(1)\right)\) \(= 4 -\left( \frac{1}{4} - \frac{2}{3}\right)\) \(= 4-\frac{1}{12}\) \(= \frac{47}{12}\) So, the value of the line integral is: \(\int_{C}x^{3}d x+y^{3}d y =\frac{47}{12}\)