Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  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 $\int \frac{dx}{a+b\sin x}$?

share|improve this question
1  
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

7 Answers 7

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

share|improve this answer

(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.

share|improve this answer

For any integral of the form $\int\frac{dx}{x(1+x^n)}$, it is effective to factor out the $x^n$ in the denominator. Therefore:

$\int\frac{dx}{x(1+x^n)}$

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

$ = -\int\frac{du}{u}$ with $u = 1+\frac{1}{x^n}$, $du = \frac{-n}{x^{n+1}}{dx}$

$ = -\frac{1}{n}\log({1+\frac{1}{x^n}})+C$

For $n=3$

$\int\frac{dx}{x(1+x^3)} = -\frac{1}{3}\log({1+\frac{1}{x^3}})+C = \log{x}-\frac{\log{(x^3+1)}}{3}+C$

share|improve this answer

There's an easier solution for 1)

$$ I = \int \frac{dx}{2 + 2\sin x + \cos x } = \int \frac {\sec x}{2(\sec x + \tan x) + 1} dx $$

Let $ u = \sec x + \tan x $

$$ du = (\sec x \tan x + \sec^2 x) \,dx = \sec x (\sec x + \tan x) \,dx $$ $$ \Rightarrow \sec x \,dx = \frac{du}{u} $$

$$ \Rightarrow I = \int \frac{du}{u(2u+1)} $$

Decompose into partial fractions and you're done

share|improve this answer

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

For 3) try splitting into partial fractions.

share|improve this answer

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

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

share|improve this answer

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 $

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.