Question

integralcos x cos 2x cos 3x dx =​

Answer 1

$\bf{\underline{\underline{To\:\:find \: : }}}$

$\mathtt{\int cosx \: cos2x \: cos3x \: dx}$

$\bf{\underline{\underline{Solution}}}$

$\mathtt{\int cosx .\: cos2x \: .cos3x \: dx}$

$\mathtt{\rightarrow\: \frac{-1}{2} \int sinx \left( -2sin2x\: . sin3x \right) dx}$

$\mathtt{\rightarrow\: \frac{-1}{2} \int sinx \left[ cos5x\: - cosx \right] dx}$

$\mathtt{\rightarrow\: \frac{-1}{2} \int sinx \: . cos5x\:dx + \frac{1}{2} \int sinx\: . cosx\:dx}$

$\mathtt{\rightarrow\: \frac{-1}{2} × \frac{1}{2} \int 2sinx \: . cos5x\:dx + \frac{1}{2} × \frac{1}{2} \int 2sinx\: . cosx\:dx}$

$\mathtt{\rightarrow\: \frac{-1}{4} \int (sin6x \: - sin4x)dx + \frac{1}{4} \int sin2x\: .dx}$

$\mathtt{\rightarrow\: \frac{-1}{4} \int sin6x \: .dx + \frac{1}{4} \int sin4x\: .dx + \frac{1}{4} \int sin2x\: .dx}$

$\mathtt{\rightarrow\: \frac{-1}{4} \: (\frac{-cos6x}{6}) + \frac{1}{4} × (\frac{-cos4x}{4}) + \frac{1}{4} × (\frac{-cos2x}{2}) + C}$

$\mathtt{\red{\rightarrow\: \frac{1}{24}\: cos6x - \frac{1}{16} \: cos4x - \frac{1}{8} \: cos2x + C}}$

$\bf{\underline{\underline{Identity\:used}}}$

$\fbox{\mathtt{\purple{\bigstar\:\:cosC - cosD = 2sin\frac{C + D}{2}\: sin\frac{C - D}{2}}}}$

$\fbox{\mathtt{\purple{\bigstar\:\:sinC - sinD = 2cos\frac{C + D}{2}\: sin\frac{C - D}{2}}}}$