NIMCET 2019 Question Paper with solutions

Question : 27 of 120

Marks: +1, -0

∫ \cos x \cos 2 x \cos 5 x d x=A_{1} \sin 2 x+A_{2} \sin 4 x+A_{3} \sin 6 x+A_{4} \sin 8 x+c

, then the values of

A_{1}

{A}_{2}, {A}_{3}, {A}_{4}

are

Solution:

Let

\I=∫\cos x\cos2x\cos5x\,dx

\I=\;\;{1}/{2}∫2\cos x\cos5x\cos2x\,dx

\I=\;\;{1}/{2}∫(\cos6x+\cos4x)\cos2x\,dx

\I=\;\;{1}/{4}∫(2\cos6x\cos2x+2\cos4x\cos2x)dx

\I=\;\;{1}/{4}∫(\cos8x+\cos4x+\cos6x+\cos2x)dx

\I=\;\;{1}/{4}[\;\;{\sin8x}/{8}+\;\;{\sin4x}/{4}+\;\;{\sin6x}/{6}+\;\;{\sin2x}/{2}]+C

\I=\;\;{1}/{8}\sin2x+\;\;{1}/{16}\sin4x+\;\;{1}/{24}\sin6x+\;\;{1}/{32}\sin8x+C

∴\;A_{1}=\;\;{1}/{8},A_{2}=\;\;{1}/{16},A_{3}=\;\;{1}/{24},A_{4}=\;\;{1}/{32}

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