Question
Integrate the following integrals:
$\int\cos3\text{x}\cos4\text{x dx}$
$\int\cos3\text{x}\cos4\text{x dx}$
✓
Answer
Let I $=\int\cos3\text{x}\cos4\text{x dx}.$ Then,
$\text{I}=\frac{1}{2}\int(2\cos3\text{x}\cos4\text{x})\text{dx}$
$=\frac{1}{2}\int(\cos7\text{x}+\cos(-\text{x}))\text{dx}$
$=\frac{1}{2}\int\cos7\text{x}+\frac{1}{2}\int\cos\text{dx}$ $[\because\cos(-0)=\cos0]$
$=\frac{\sin7\text{x}}{2\times7}+\frac{\sin\text{x}}{2}+\text{C}$
$=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\sin\text{x}+\text{C}$
$\therefore\text{I}=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\times\sin\text{x}+\text{C}$
$\text{I}=\frac{1}{2}\int(2\cos3\text{x}\cos4\text{x})\text{dx}$
$=\frac{1}{2}\int(\cos7\text{x}+\cos(-\text{x}))\text{dx}$
$=\frac{1}{2}\int\cos7\text{x}+\frac{1}{2}\int\cos\text{dx}$ $[\because\cos(-0)=\cos0]$
$=\frac{\sin7\text{x}}{2\times7}+\frac{\sin\text{x}}{2}+\text{C}$
$=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\sin\text{x}+\text{C}$
$\therefore\text{I}=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\times\sin\text{x}+\text{C}$
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