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Integrate: $$\int \frac{\sin(x)}{9+16\sin(2x)}\,\text{d}x.$$

I tried the substitution method ($\sin(x) = t$) and ended up getting $\int \frac{t}{9+32t-32t^3}\,\text{d}t$. Don't know how to proceed further.

Also tried adding and substracting $\cos(x)$ in the numerator which led me to get $$\sin(2x) = t^2-1$$ by taking $\sin(x)+\cos(x) = t$.

Can't figure out any other method now. Any suggestions or tips?

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  • $\begingroup$ Try the angent half-angle substitution: en.wikipedia.org/wiki/Tangent_half-angle_substitution. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 21 '16 at 18:22
  • $\begingroup$ And in the substitution $\sin x = t$ you've fogotten the $dx$. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 21 '16 at 18:23
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    $\begingroup$ Is this integral given this way, or as a defined integral? Because if you need to integrate from 0 to $2 \pi$ that is an easier task. $\endgroup$ – N. S. Jan 21 '16 at 18:50
  • $\begingroup$ @N.S. I was given in this way but I would love to see the method on how to solve this integral from 0 to 2pi. $\endgroup$ – Gauz Jan 21 '16 at 18:59
  • $\begingroup$ @Gauz If you use $z(t) =cos(t)+i \sin(t)$ you can express this integral as the integral of a rational function over the circle of radius one in the complex plane. Then the residue Theorem calculates this immediately. The solution relies on complex analysis though... $\endgroup$ – N. S. Jan 21 '16 at 19:01
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To attack this integral, we will need to make use of the following facts:

$$(\sin{x} + \cos{x})^2 = 1+\sin{2x}$$

$$(\sin{x} - \cos{x})^2 = 1-\sin{2x}$$

$$\text{d}(\sin{x}+\cos{x}) = (\cos{x}-\sin{x})\text{d}x$$

$$\text{d}(\sin{x}-\cos{x}) = (\cos{x}+\sin{x})\text{d}x$$

Now, consider the denominator.

It can be rewritten in two different ways as hinted by the above information.

$$9+16\sin{2x} = 25 - 16(1-\sin{2x}) = 16(1+\sin{2x})-7$$

$$9+16\sin{2x} = 25 - 16(\sin{x}-\cos{x})^2 = 16(\sin{x}+\cos{x})^2-7$$

Also note that

$$\text{d}(\sin{x}-\cos{x})-\text{d}(\sin{x}+\cos{x}) = 2\sin{x}\text{d}x$$

By making the substitutions

$$u = \sin{x}+\cos{x}, v = \sin{x}-\cos{x}$$

The integral is transformed into two separate integrals which can be evaluated independently.

$$2I = \int \frac{\text{d}v - \text{d}u}{9+16\sin{2x}} = \int \frac{\text{d}v}{25-16v^2} + \int \frac{\text{d}u}{7-16u^2}$$

The remainder of this evaluation is left as an exercise to the reader.

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HINT:

$$\int\frac{\sin(x)}{9+16\sin(2x)}\space\text{d}x=$$


Use the double angle formula $\sin(2x)=2\sin(x)\cos(x)$:


$$\int\frac{\sin(x)}{32\sin(x)\cos(x)+9}\space\text{d}x=$$


Subsitute $u=\tan\left(\frac{x}{2}\right)$ and $\text{d}u=\frac{\sec^2\left(\frac{x}{2}\right)}{2}\space\text{d}x$.

Then transform the integrand using the substitutions:

$\sin(x)=\frac{2u}{u^2+1},\cos(x)=\frac{1-u^2}{u^2+1}$ and $\text{d}x=\frac{2}{u^2+1}\space\text{d}u$:


$$\int\frac{4u}{\left(u^2+1\right)^2\left(\frac{64u(1-u^2)}{(u^2+1)^2}+9\right)}\space\text{d}u=$$ $$\int\frac{4u}{9 u^4-64 u^3+18 u^2+64 u+9}\space\text{d}u$$

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