# Integrals from MIT integration bee

1. $\int\frac{dx}{2+2\sin x+\cos x}$
2. $\int_0^{\infty}\frac{\ln x}{1+x^2}dx$
3. $\int\frac{dx}{x(1+x^3)}$

In general what is $\int \frac{dx}{a+b\sin x}$?

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try to use $\sin(x) = \dfrac{2\tan(x/2)}{1+\tan^2(x/2)}$. –  Yimin Feb 20 '13 at 4:28
To do (2), you can integrate $\frac{(\ln x)^2}{1+x^2}$ on a Pacman contour, as in this example. –  Micah Feb 20 '13 at 4:40
See this. –  Ragib Zaman Feb 20 '13 at 4:42

For $2$, $$\int_0^{\infty} \dfrac{\ln(x)}{1+x^2} dx = \int_0^1 \dfrac{\ln(x)}{1+x^2} dx + \int_1^{\infty} \dfrac{\ln(x)}{1+x^2} dx$$ $$\int_1^{\infty} \dfrac{\ln(x)}{1+x^2} dx = -\int_1^0 \dfrac{\ln(1/x)}{1+1/x^2} \dfrac{dx}{x^2} = -\int_0^1 \dfrac{\ln(x)}{1+x^2} dx$$ Hence, $$\int_0^{\infty} \dfrac{\ln(x)}{1+x^2} dx = \int_0^1 \dfrac{\ln(x)}{1+x^2} dx + \int_1^{\infty} \dfrac{\ln(x)}{1+x^2} dx = 0$$

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(1) Use Weierstrass substitution

(2) Putting $x=\tan\theta$

$$I=\int_0^{\infty}\frac{\ln x}{1+x^2}dx=\int_0^{\frac\pi2}\log\tan\theta d\theta=\int_0^{\frac\pi2}\log\tan\left(\frac\pi2+0-\theta\right)d\theta$$ as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

$$I=\int_0^{\frac\pi2}\log\cot\theta d\theta=-\int_0^{\frac\pi2}\log\tan\theta d\theta=-I\implies I=0$$

(3) $$\int\frac{dx}{x(1+x^3)}=\int\frac{x^2dx}{x^3(1+x^3)}$$

Put $x^3=y$ and then apply Partial Fraction Decomposition rule.

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Hint: For 2) substitute $x= \tan\theta$

For 3) try splitting into partial fractions.

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For the first and the general ones: trigonometric substitution:

$$u=\tan\frac{x}{2}\Longrightarrow dx=\frac{2\;du}{1+u^2}\,\,,\,\,\sin x=\frac{2u}{1+u^2}\,\,,\,\,\cos x=\frac{1-u^2}{1+u^2}$$

For the last one:

$$x^3+1=(x+1)(x^2-x+1)$$

and then by partial fractions

$$\frac{1}{x(x^3+1)}=\frac{1}{x(x+1)(x^2-x+1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{x^2-x+1}\ldots etc.$$

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Nothing for 3) yet, funny.

Solution 3) Split up into partial fractions just as Ishan Banerjee suggested. The integrand is $$\dfrac {1}{x(1+x^3)} = \dfrac {1}{x(1+x)(1-x+x^2)} = \dfrac {A}{x} + \dfrac {B}{x+1} + \dfrac {C}{x^2-x+1}.$$

The RHS can be added out: \begin {align*} \dfrac {A(x+1) + Bx}{x(x+1)} + \dfrac {C}{x^2 - x + 1} &= \dfrac {A(x+1)(x^2-x+1) + Bx(x^2-x+1) + Cx(x+1)}{x(1+x^3)} \\&= \dfrac {(A+B)x^3 + (C-B)x^2 + (B+C)x + A}{x(1+x^3)} \end {align*}

$\implies (A+B)x^3 + (C-B)x^2 + (B+C)x + A = 1$

$\implies \left\{\begin{array}{lr} A + B = 0, \\ C - B = 0, \\ B + C = 0, \\ A = 1. \end{array}\right.$

Obviously, we have a contradiction, which means that this partial fractions representation is incorrect. This is often times a key technique to checking what partial fractions representation you want, if you are unsure. In the Integration Bee, one should be more experienced and recognize that it must be of the form $$\dfrac {A}{x} + \dfrac {B}{x+1} + \dfrac {Cx+D}{x^2-x+1}.$$You can solve it the same way, and I will leave that as an exercise.

Anyway, we get that $$\dfrac {1}{x(1+x^3)} = \dfrac {1}{x} - \dfrac {1}{3(x+1)} - \dfrac {2x-1}{3(x^2-x+1)}.$$

Now, we integrate: $$\displaystyle\int \dfrac {\mathrm{d}x}{x(1+x^3)} = \displaystyle\int \dfrac {1}{x} \, \mathrm{d}x - \displaystyle\int \dfrac {1}{3(x+1)} \, \mathrm{d}x - \displaystyle\int \dfrac {2x-1}{3(x^2-x+1)} \, \mathrm{d}x.$$

Integrating these, we get our final answer to be $$\boxed {\log x - \dfrac {\log (x^3+1)}{3} + \mathcal{C}}.$$

$\blacksquare$

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