Question

Integrate 2sin 2x -cosx/ 6 -cos^2x -4sinx​

Answer 1

Using the identity

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

we have

$\dfrac{2 \sin(2x) - \cos(x)}{6 - \frac{1 + \cos(2x)}2 - 4 \sin(x)} = \dfrac{4 \sin(2x) - 2 \cos(x)}{11 - \cos(2x) - 8 \sin(x)} \\\\ ~~~~~~~~ = \dfrac{2(2 \sin(2x) - 8 \cos(x)) + 14 \cos(x)}{11 - \cos(2x) - 8 \sin(x)}$

Expand the integral as

$\displaystyle \int \frac{2 \sin(2x) - \cos(x)}{6 - \cos^2(x) - 4 \sin(x)} \, dx \\\\ ~~~~= 2 \int \frac{2 \sin(2x) - 8 \cos(x)}{11 - \cos(2x) - 8 \sin(x)} \, dx + 14 \int \frac{\cos(x)}{11 - \cos(2x) - 8 \sin(x)} \, dx$

In the first integral, substitute

$y = 11 - \cos(2x) - 8 \sin(x) \implies dy = \bigg(2\sin(2x) - 8 \cos(x) \bigg) \, dx$

In the second integral, rewrite the denominator in terms of $\sin(x)$.

$11 - \cos(2x) - 8\sin(x) = 11 - (1 - 2\sin^2(x)) - 8\sin(x) \\\\ ~~~~~~~~ = 10 - 8\sin(x) + 2 \sin^2(x)$

Now substitute

$z = \sin(x) \implies dz = \cos(x) \, dx$

and complete the square.

$2z^2 - 8z + 10 = 2 (z-2)^2 + 2$

Then we have

$\displaystyle \int \frac{2 \sin(2x) - \cos(x)}{6 - \cos^2(x) - 4 \sin(x)} \, dx = 2 \int \frac{dy}y + 7 \int \frac{dz}{(z-2)^2 + 1}$

In the $z$-integral, substitute

$w = z-2 \implies dw = dz$

Then the integral is

$\displaystyle \int \frac{2 \sin(2x) - \cos(x)}{6 - \cos^2(x) - 4 \sin(x)} \, dx = 2 \int \frac{dy}y + 7\int \frac{dw}{w^2 + 1} \\\\ ~~~~~~~~ = 2 \ln|y| + 7 \tan^{-1}(w) + C \\\\ ~~~~~~~~ = 2\ln|y| + 7\tan^{-1}(z - 2) + C \\\\ ~~~~~~~~ = \boxed{2\ln(11 - \cos(2x) - 8 \sin(x)) + 7 \tan^{-1}(\sin(x) - 2) + C}$