Mathematics

# Solve that :$\int {\sin 4x.\cos 3x}$ dx

##### SOLUTION
$\displaystyle\int{\sin{4x}\cos{3x}dx}$

$=\dfrac{1}{2}\displaystyle\int{2\sin{4x}\cos{3x}dx}$

$=\dfrac{1}{2}\displaystyle\int{\left(\sin{7x}+\sin{x}\right)dx}$ ........  using  $2\sin{A}\cos{B}=\sin{\left(A+B\right)}+\sin{\left(A-B\right)}$

$=\dfrac{1}{2}\left[\dfrac{-\cos{7x}}{7}-\cos{x}\right]+c$

$=-\dfrac{1}{14}\cos{7x}-\dfrac{1}{2}\cos{x}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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• A. $\displaystyle \frac{\pi}{4}$
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• C. $\displaystyle \frac{\pi}{6}$
• D. $\displaystyle \frac{\pi}{12}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Integrate with respect to $x$:
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1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate the given integral.
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Q4 Single Correct Medium
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