Mathematics

# Evaluate :$\int x \sin 3x \ dx$

$-\dfrac{1}{3}x \cos 3x+\dfrac{1}{9} \sin 3x +C$

##### SOLUTION
$I=\int x \sin 3x \ dx$
$I=x[\int \sin 3x \ dx]-\int [\dfrac{d}{dx}(x) \times \int \sin 3x \ dx]dx$

$I=x \times -\dfrac{1}{3} \cos 3x -\int [-\dfrac{1}{3} \cos 3x] dx$

$I=-\dfrac{1}{3}x \cos 3x +\dfrac{1}{3} \int \cos 3x \ dx =-\dfrac{1}{3}x\cos 3x +\dfrac{1}{9} \sin 3x +C$

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Integrate $\displaystyle \int \frac{x}{x^2+1}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int { \sin ^{ 5 }{ x }\ dx= }$
• A. $\dfrac { -\sin ^{ 4 }{ x } \cos { x } }{ 5 } -\dfrac { 4 }{ 15 } \sin ^{ 2 }{ x } \cos { x+\dfrac { 8 }{ 15 } \cos { x } +c }$
• B. $\dfrac { -\sin ^{ 4 }{ x } \cos { x } }{ 5 } +\dfrac { 4 }{ 15 } \sin ^{ 2 }{ x } \cos { x+\dfrac { 8 }{ 15 } \cos { x } +c }$
• C. $none$
• D. $\dfrac { -\sin ^{ 4 }{ x } \cos { x } }{ 5 } -\dfrac { 4 }{ 15 } \sin ^{ 2 }{ x } \cos { x-\dfrac { 8 }{ 15 } \cos { x } +c }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
f $k = e^{2007}$ then value of $\displaystyle I =\int_{1}^{k}\frac{ \pi \cos (\pi \log x )} {x} dx$ is
• A. $-\pi$
• B. $\pi/e$
• C. $2007\pi$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
$\displaystyle \int{\dfrac{dx}{x^{\dfrac{1}{2}}+x^{\dfrac{1}{3}}}}$

1 Verified Answer | Published on 17th 09, 2020

Q5 Subjective Medium
Solve $\displaystyle \int \cos^{3}x\ dx$