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Is there easier way than universal supstitution to solve this integral $$\int\frac{2+\sin x}{\sin x(1+\cos x)}dx$$?

Meow
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4 Answers4

5

The integral $\int\frac{2}{\sin x(1+\cos x)}\,dx$ can be found by multiplying top and bottom by $\sin x$. After replacing the $\sin^2 x$ in the denominator by $1-\cos ^2 x$, we make the substitution $1-\cos x=u$. We end up with an integration that can be done using partial fractions.

For the remaining integral $\int \frac{1}{1+\cos x}\,dx$, use the fact this is $\int\frac{1}{2\cos^2(x/2)}\,dx$, so we want to integrate $\frac{1}{2}\sec^2 (x/2)$, easy.

user84413
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André Nicolas
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First noting that $\frac{1+\cos(x)}{2}=\cos^2(x/2)$ and then using partial fractions,

$$ \begin{align} &\int\frac{2+\sin(x)}{\sin(x)(1+\cos(x))}\mathrm{d}x\\ &=\int\frac{2\,\mathrm{d}x}{\sin(x)(1+\cos(x))}+\int\frac{\mathrm{d}x}{1+\cos(x)}\\ &=\int\frac{-2\,\mathrm{d}\cos(x)}{(1-\cos^2(x))(1+\cos(x))}+\int\sec^2(x/2)\,\mathrm{d}x/2\\ &=-2\int\left(\frac{1/4}{1-\cos(x)}+\frac{1/4}{1+\cos(x)}+\frac{1/2}{(1+\cos(x))^2}\right)\mathrm{d}\cos(x)+\tan(x/2)\\ &=\frac12\log\left(\frac{1-\cos(x)}{1+\cos(x)}\right)+\frac1{1+\cos(x)}+\frac{\sin(x)}{1+\cos(x)}+C\\ &=\log\left(\frac{\sin(x)}{1+\cos(x)}\right)+\frac{1+\sin(x)}{1+\cos(x)}+C \end{align} $$

Disclaimer: Having finally read André's answer, I see that my answer closely follows his outline. So consider this an implementation of his program.

robjohn
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  • Unrelated to the mathematical content of your answer, but are we supposed to write the differential using "\mathrm{d}x" instead of "dx"? I've been using just regular "dx" which produces the italicized "d" in the output. Have I been wrong the whole time? :O – Cookie Aug 01 '14 at 06:46
  • It is purely stylistic. I believe that most typesetters will set the $\mathrm{d}$ in a font or style distinct from the variables so that it is not mistaken for a variable. Often it is clear from the context, so not changing font or style is not wrong, but I prefer to set the $\mathrm{d}$ apart. I think it also emphasizes that $\int f(x),\mathrm{d}x$ is operating on $f(x)$ and $x$. – robjohn Aug 01 '14 at 06:56
1

We can rewrite the integral as $\displaystyle\int\frac{2}{\sin x(1+\cos x)} dx +\int \frac{1}{1+\cos x} dx$.

In the first integral, multiplying by $\sin x$ on the top and bottom gives $\displaystyle\int\frac {2\sin x}{(1-\cos^{2}x)(1+\cos x)} dx = \int\frac{-2}{(1+u)^2(1-u)} du$ after substituting $u=\cos x$,

and then partial fractions can be used to get

$\displaystyle -\int\bigg(\frac{1/2}{1+u}+\frac{1}{(1+u)^2}+\frac{1/2}{1-u}\bigg) du=-\bigg(\frac{1}{2}\ln(1+u)-(1+u)^{-1}-\frac{1}{2}\ln(1-u)\bigg) du$

$\displaystyle=-\frac{1}{2}\ln(1+\cos x)+(1+\cos x)^{-1}+\frac{1}{2}\ln(1-\cos x)+C$

$\displaystyle =\ln\vert\csc x-\cot x\vert+(1+\cos x)^{-1}+C$.

In the second integral, multiplying by $1-\cos x$ on the top and bottom gives

$\displaystyle\int\frac{1-\cos x}{\sin^{2}x} dx=\int(\csc^{2}x-\csc x\cot x)\; dx = -\cot x+\csc x+C$.

user84413
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0

Well, first step I see would be to multiply numerator and denominator by $1-\cos x$.

$$\int\dfrac{(1-\cos x)(2+\sin x)dx}{\sin^3x}=\int2\csc^3x+\csc^2x-2\csc^2 x\cot x-\csc x\cot xdx$$

The last 3 terms integrate quickly and easily and I'm guessing you've already seen how to handle $\int\csc^3xdx$.

Mike
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