Mathematics

# Solve $\int {\cos 4x\cos xdx}$

##### SOLUTION

Consider the given integral.

$I=\int{\left( \cos 4x\cos x \right)dx}$

We know that

$\cos A\cos B=\dfrac{1}{2}\left[ \cos \left( A+B \right)+\cos \left( A-B \right) \right]$

Therefore,

$I=\int{\dfrac{1}{2}\left[ \cos \left( 4x+x \right)+\cos \left( 4x-x \right) \right]dx}$

$I=\dfrac{1}{2}\int{\left[ \cos \left( 5x \right)+\cos \left( 3x \right) \right]dx}$

$I=\dfrac{1}{2}\left[ \dfrac{\sin 5x}{5}+\dfrac{\sin 3x}{3} \right]+C$

$I=\dfrac{\sin 5x}{10}+\dfrac{\sin 3x}{6}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 128

#### Realted Questions

Q1 Single Correct Hard
The value of $\int _{ 0 }^{ 1 }{ { x }^{ 2 }{ \left( 1-x \right) }^{ 9 }dx }$ is
• A. $\dfrac { 1 }{ 610 }$
• B. $\dfrac { 1 }{ 630 }$
• C. $\dfrac { 1 }{ 640 }$
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1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Solve: $\displaystyle \int { \dfrac { 1 }{ 4{ x }^{ 2 }+4x+5 } dx }$
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• C. None of these
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1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
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