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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

juantheron
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  • Hint: For integral $(b)$, consider change of variables

    $$y = \sqrt{1+x^3}-1 \quad\iff\quad x = \sqrt[3]{(y+1)^2-1} = \sqrt[3]{y^2 + 2y}$$

    for the part $\int_0^2 \sqrt{1+x^3} dx$ and then combine the result with the other part...

     – achille hui Oct 22 '14 at 05:24
  • $$\int_0^1\dfrac{1+t^k}{(1+t)^{k+2}}dt=\dfrac1{k+1}$$ Try $u=1+t$ and expand $t^k=(u-1)^k$ using the binomial formula, then switch the order of summation and integration. – Lucian Oct 22 '14 at 10:52
  • @Lucian the second integral is $6$. – achille hui Oct 22 '14 at 11:51
  • @achillehui: I stand corrected! :-$)$ – Lucian Oct 22 '14 at 12:01
  • Thanks achille hui and Lucian. but I did not understand how can I Get $\displaystyle \int_{0}^{1}\frac{1+t^{k}}{(1+t)^{k+2}}dt = \frac{1}{k+1}$ I have used yours Hint and get $\displaystyle \int_{1}^{2}u^{-(k+2)}du+\int_{1}^{2}\left(1-\frac{1}{u}\right)\cdot \frac{1}{u^2}du$ – juantheron Oct 23 '14 at 02:32

2 Answers2

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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.

Claude Leibovici
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3

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,

$$\int_0^2 \left(\sqrt{1+x^3} + \sqrt[3]{x^2+2x}\right) dx = \int_0^2 \left( 1 + \sqrt[3]{x^2+2x} \right) dx + \int_0^2 y\, d\sqrt[3]{y^2+2y}\\ = \left[ x + x\sqrt[3]{x^2+2x}\right]_0^2 = 2 + 2\sqrt[3]{2^2+2\cdot 2} = 6 $$

achille hui
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