Answer:
[tex]\int\limits^._c {(y+5sinx)} \, dx +(z^{2}+2cosy )dy+x^{3} dz = \pi[/tex]
Step-by-step explanation:
The line integral is [tex]\int\limits^._c {(y+5sinx)} \, dx +(z^{2}+2cosy )dy+x^{3} dz\\ C:r(t)=sint,cost,sin2t, where , 0$\leq$t$\leq$2\pi\\x=sint,y=cost, z=sin2t\\dx=costdt,dy=-sintdt,dz=2cos2tdt\\therefore[/tex]
[tex]\int\limits^._c {(y+5sinx)} \, dx +(z^{2}+2cosy )dy+x^{3} dz[/tex]
[tex]\int\limits^._c {(cost+5sin(sint)} \, costdt +(sin^{2}2t+2cos(cost) ).-sintdt+sin^{3}t(2cos2tdt)\\[/tex]
[tex]$\int_{0}^{2\pi} [{(cos^{2} t+5costsin(sint)} \, -(sintsin^{2}2t+2sintcos(cost) )+2sin^{3}t(cos2t)]dt\\[/tex]
Now evaluate the integrals separately
[tex]=$\int_{0}^{2\pi} cos^2t dt$ =$\int_{0}^{2\pi}\frac{1+cos2t}{2}dt=\frac{1}{2}$\int_{0}^{2\pi}(1+cos2t)dt[/tex]
[tex]=\frac{1}{2}[t+\frac{1}{2}sin2t] =\frac{1}{2}2\pi=\pi[/tex]
[tex]$\int_{0}^{2\pi}5costsin(sint)dt=0\\[/tex]
[tex]$\int_{0}^{2\pi}sintsin^{2}2tdt[/tex]
[tex]=$\int_{0}^{2\pi}sint(2sintcost)^{2}dt[/tex]
[tex]4$\int_{0}^{2\pi}sin^{3}tcos^{2}tdt = 0 ,by,maple[/tex]
[tex]$\int_{0}^{2\pi}2sintcos(cost)dt = 0 ,by,maple[/tex]
[tex]$\int_{0}^{2\pi}2sin^{3}t(cos2t)dt = 0 ,by,maple[/tex]
therefore
[tex]\int\limits^._c {(y+5sinx)} \, dx +(z^{2}+2cosy )dy+x^{3} dz = \pi[/tex]
