Evaluation of $\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx$ and $\int_{0}^{2} \left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\right)dx$ Evaluation of Some Integrals::
$\displaystyle (a)\;\;\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\sin 3x}{1+2\sin x}\right)dx\;\;\;\;\;\;(b)\;\; \int_{0}^{2} \left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\right)dx\;\;\;\;\;\;$
$\displaystyle (c)\;\;\int_{0}^{1}\frac{4x^3\cdot \left(1+(x^4)^{2010}\right)}{\left(1+x^4\right)^{2012}}dx\;\;\;\;\;\; (d)\;\; $
$\bf{My\; Try::}$I have tried for third one:: Let $\displaystyle I = \int_{0}^{1}\frac{4x^3\cdot \left(1+(x^4)^{2010}\right)}{\left(1+x^4\right)^{2012}}dx$
Now Put $\displaystyle x^4=t\;,$ Then $\displaystyle 4x^3dx=dt$ and changing Limit, we Get
$\displaystyle I = \int_{0}^{1}\frac{1+t^{2010}}{(1+t)^{2012}}dt$
Now How can I solve after that , Help me
and i did not understand how can i solve $(a)$ and $(b)\;,$ Help me
Thanks

Hint
For the first integral, use $\sin(3x)=3\sin(x)-4\sin^3(x)$ . So $$\frac{1+\sin 3x}{1+2\sin x}=\frac{-4 \sin(x)^3+3 \sin(x)+1}{2 \sin(x)+1}$$ Remark that equation $-4z^3+3z+1=0$ has an obvious solution $z=1$ and two other roots equal to $z=-\frac{1}{2}$ ... which is also the root of $2z+1=0$ ! 
Then,
$$\frac{1+\sin 3x}{1+2\sin x}=\frac{-4 \sin(x)^3+3 \sin(x)+1}{2 \sin(x)+1}=-2 \sin^2(x)+\sin(x)+1=\cos(2x)+\sin(x)$$
I am sure that you can take from here.
