Compute the following indefinite integrals:
- \(\displaystyle \int \cos\!\left(x+\frac\pi4\right)\cos\!\left(x-\frac\pi4\right)\,dx\).
- \(\displaystyle \int \sin(4x)\cos(3x)\,dx\).
- \(\displaystyle \int 4\cos x\cos(2x)\sin(3x)\,dx\).
Solution
1. Use product-to-sum.
\[ \cos\!\left(x+\frac\pi4\right)\cos\!\left(x-\frac\pi4\right) =\frac12\left(\cos(2x)+\cos\frac\pi2\right) =\frac12\cos(2x). \] Hence \[ \boxed{\int \cos\!\left(x+\frac\pi4\right)\cos\!\left(x-\frac\pi4\right)\,dx =\frac14\sin(2x)+C.} \]
2. Use product-to-sum again.
\[ \sin(4x)\cos(3x)=\frac12\bigl(\sin(7x)+\sin x\bigr). \] Therefore \[ \boxed{\int \sin(4x)\cos(3x)\,dx =-\frac1{14}\cos(7x)-\frac12\cos x+C.} \]
3. Reduce the triple product.
Since \[ 2\cos x\cos(2x)=\cos(3x)+\cos x, \] we have \[ \begin{aligned} 4\cos x\cos(2x)\sin(3x) &=2\sin(3x)\bigl(\cos(3x)+\cos x\bigr)\\ &=\sin(6x)+\sin(4x)+\sin(2x). \end{aligned} \] Thus \[ \boxed{\int4\cos x\cos(2x)\sin(3x)\,dx =-\frac16\cos(6x)-\frac14\cos(4x)-\frac12\cos(2x)+C.} \]