<s>[INST] 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
 [/INST] Despite its appearance, there is actually a surprisely simple way to evaluate integral $(b)$. 
Consider following change of variables,
$$y = \sqrt{1+x^3}-1 \quad\iff\quad x = \sqrt[3]{(y+1)^2-1} = \sqrt[3]{y^2 + 2y}$$
When $x$ varies from $0$ to $2$, the corresponding $y$ varies from $0$ to $2$ too.
As a result,
$$\int_0^2 \left(\sqrt{1+x^3} + \sqrt[3]{x^2+2x}\right) dx
= \int_0^2 \left( 1 + \sqrt[3]{x^2+2x} \right) dx + \int_0^2 y\, d\sqrt[3]{y^2+2y}\\
= \left[ x + x\sqrt[3]{x^2+2x}\right]_0^2
= 2 + 2\sqrt[3]{2^2+2\cdot 2} = 6
$$
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