∫(cos 5x+cos 4x)/(1-2cos 3x) dx is equal to

The correct answer is Option (3) → \(\displaystyle -\frac{1}{2}\sin 2x-\sin x+C\)

The integrand simplifies once everything is written in terms of the half angles. By the sum-to-product formula:

    \( \displaystyle \cos 5x+\cos 4x = 2\cos\frac{9x}{2}\cos\frac{x}{2} \)

Using \(\cos 3x = 2\cos^2\frac{3x}{2}-1\) in the denominator:

    \( \displaystyle 1-2\cos 3x = 1-4\cos^2\frac{3x}{2}+2 \)

    \( \displaystyle = 3-4\cos^2\frac{3x}{2} \)

Since \(\frac{9x}{2} = 3\times\frac{3x}{2}\), the triple-angle formula \(\cos 3\theta = 4\cos^3\theta-3\cos\theta\) gives

    \( \displaystyle \cos\frac{9x}{2} = 4\cos^3\frac{3x}{2}-3\cos\frac{3x}{2} \)

    \( \displaystyle = -\cos\frac{3x}{2}\left(3-4\cos^2\frac{3x}{2}\right) \)

So the bracket cancels with the denominator:

    \( \displaystyle \frac{\cos 5x+\cos 4x}{1-2\cos 3x} = -2\cos\frac{3x}{2}\cos\frac{x}{2} \)

Using \(2\cos A\cos B = \cos(A+B)+\cos(A-B)\):

    \(-2\cos\frac{3x}{2}\cos\frac{x}{2} = -(\cos 2x+\cos x)\)

Integrating:

    \( \displaystyle \int-(\cos 2x+\cos x)\,dx = -\frac{1}{2}\sin 2x-\sin x+C \)

Shortcut: Differentiate the options and compare with the integrand at x = 0, where the integrand is \(\frac{1+1}{1-2} = -2\):

    Option (1): \(\cos 2x-\cos x\) gives 0
    Option (2): \(\sin 2x+\sin x\) gives 0
    Option (3): \(-\cos 2x-\cos x\) gives −2
    Option (4): \(-3\cos 3x-\cos x\) gives −4

Only Option (3) agrees.