Question

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

Answer 1

Using the identity

`\cos^2(x) = \frac{1+\cos(2x)}2`

we have

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

Expand the integral as

`\displaystyle \int (2 \sin(2x) - \cos(x))/(6 - \cos^2(x) - 4 \sin(x)) \, dx \\\\ ~~~~= 2 \int (2 \sin(2x) - 8 \cos(x))/(11 - \cos(2x) - 8 \sin(x)) \, dx + 14 \int (\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 (2 \sin(2x) - \cos(x))/(6 - \cos^2(x) - 4 \sin(x)) \, dx = 2 \int \frac{dy}y + 7 \int (dz)/((z-2)^2 + 1)`

In the
`z`-integral, substitute

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

Then the integral is

`\displaystyle \int (2 \sin(2x) - \cos(x))/(6 - \cos^2(x) - 4 \sin(x)) \, dx = 2 \int \frac{dy}y + 7\int (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}`