Answer :
We can compute the integral directly: we have
[tex]\begin{cases}x(t)=\sin t\\y(t)=\cos t\\z(t)=\sin(2t)\end{cases}\implies\begin{cases}\mathrm dx=\cos t\,\mathrm dt\\\mathrm dy=-\sin t\,\mathrm dt\\\mathrm dz=2\cos(2t)\,\mathrm dt\end{cases}[/tex]
Then the integral is
[tex]\displaystyle\int_C(y+9\sin x)\,\mathrm dx+(z^2+4\cos y)\,\mathrm dy+x^3\,\mathrm dz[/tex]
[tex]=\displaystyle\int_0^{2\pi}\bigg((\cos t+9\sin(\sin t))\cos t-(\sin^2(2t)+4\cos(\cos t))\sin t+\sin^3t\bigg)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}\cos^2t+9\cos t\sin(\sin t)-\sin^2(2t)\sin t-4\sin t\cos(\cos t)+\sin^3t\,\mathrm dt[/tex]
You could also take advantage of Stokes' theorem, which says the line integral of a vector field [tex]\vec F[/tex] along a closed curve [tex]C[/tex] is equal to the surface integral of the curl of [tex]\vec F[/tex] over any surface [tex]S[/tex] that has [tex]C[/tex] as its boundary.
In this case, the underlying field is
[tex]\vec F(x,y,z)=\langle y+9\sin x,z^2+4\cos y,x^3\rangle[/tex]
which has curl
[tex]\mathrm{curl}\vec F(x,y,z)=-\langle2z,3x^2,1\rangle[/tex]
We can parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=\langle u\cos v,u\sin v,2u^2\cos v\sin v\rangle=\langle u\cos v,u\sin v,u^2\sin(2v)\rangle[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex].
Note that when viewed from above, [tex]C[/tex] has negative orientation (a particle traveling on this path moves in a clockwise direction). Take the normal vector to [tex]S[/tex] to be pointing downward, given by
[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle2u^2\sin v,2u^2\cos v,-u\rangle[/tex]
Then the integral is
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S\mathrm{curl}\vec F(x,y,z)\cdot\mathrm d\vec S[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^1-\langle2u^2\sin(2v),3u^2\cos^2v,1\rangle\cdot\langle2u^2\sin v,2u^2\cos v,-u\rangle\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle-\int_0^{2\pi}\int_0^14u^4\sin(2v)\sin v+6u^4\cos^3v-u\,\mathrm du\,\mathrm dv[/tex]
Both integrals are kind of tedious to compute, but personally I prefer the latter method. Either way, you end up with a value of [tex]\boxed\pi[/tex].