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

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?

• Try the angent half-angle substitution: en.wikipedia.org/wiki/Tangent_half-angle_substitution. Commented Jan 21, 2016 at 18:22
• And in the substitution $\sin x = t$ you've fogotten the $dx$. Commented Jan 21, 2016 at 18:23
• 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. Commented Jan 21, 2016 at 18:50
• @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. Commented Jan 21, 2016 at 18:59
• @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... Commented Jan 21, 2016 at 19:01

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.

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 $\text{d}x=\frac{2}{u^2+1}\space\text{d}u$:

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