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

Integrate 2sin 2x -cosx/ 6 -cos^2x -4sinx​-example-1
User Astaroth
by
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1 Answer

21 votes
21 votes

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}

User Jejuni
by
2.8k points
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