Question

If \[ \int \frac{\cos\left(\frac{7x}{2}\right)}{\cos\left(\frac{x}{2}\right)}\,dx = a\sin x+b\sin2x+c\sin^3x-dx+k, \] then \(a+b+c+d=\)

• A. \(1\)
• B. \(-\dfrac23\)
• C. \(\dfrac43\)
• D. \(2\)

Hint:

Use the identity \[ \boxed{\frac{\cos7A}{\cos A} = 64\cos^6A-112\cos^4A+56\cos^2A-7} \] and then convert powers of \[ \cos\frac{x}{2} \] into multiple-angle expressions before integrating.

Answer 1

The Correct Option is C

Step 1: Simplify the integrand. Using \[ \frac{\cos7A}{\cos A} = 64\cos^6A-112\cos^4A+56\cos^2A-7, \] with \[ A=\frac{x}{2}, \] and expressing powers of \[ \cos\frac{x}{2} \] in terms of \[ \cos x, \] we obtain \[ \frac{\cos\left(\frac{7x}{2}\right)} {\cos\left(\frac{x}{2}\right)} = 4\cos3x+4\cos2x+\cos x-1. \]

Step 2: Integrate each term. Therefore, \[ \int\frac{\cos\left(\frac{7x}{2}\right)} {\cos\left(\frac{x}{2}\right)}\,dx = \frac43\sin3x +2\sin2x +\sin x -x+k. \] Using \[ \sin3x=3\sin x-4\sin^3x, \] we get \[ \frac43\sin3x = 4\sin x-\frac{16}{3}\sin^3x. \] Hence, \[ \int\frac{\cos\left(\frac{7x}{2}\right)} {\cos\left(\frac{x}{2}\right)}\,dx = 5\sin x +2\sin2x -\frac{16}{3}\sin^3x -x+k. \] Comparing with \[ a\sin x+b\sin2x+c\sin^3x-dx+k, \] we have \[ a=5,\qquad b=2,\qquad c=-\frac{16}{3},\qquad d=1. \]

Step 3: Find the required sum. Therefore, \[ a+b+c+d = 5+2-\frac{16}{3}+1 = \frac43. \] Hence, \[ \boxed{\frac43}. \] Thus, \[ \boxed{(C)} \] is the correct answer.