Question

If $\int (3x^2+x-2){\sin }^2(3x+1)dx$
$=-\frac{u\left(x\right)}{72}\sin (6x+2)+\frac{v\left(x\right)}{72}\cos (6x+2)+\frac{1}{2}{x}^3+\frac{1}{4}{x}^2-x+C$

• A. $u\left(x\right)=3x^2+6x-13$
• B. $u\left(x\right)=18x^2+2x-13$
• C. $v\left(x\right)=3x+1$
• D. $v\left(x\right)=3-(6x+1)$

Answer 1

The correct option is D $v\left(x\right)=3-(6x+1)$

Let $I=\int (3x^2+x-2){\sin }^2(3x+1)dx=\frac{1}{2}\int (3x^2+x-2)(1-\cos (6x+2)).$
Now applying the formula in comprehension, the last integral can be written as
$\frac{1}{2}[x^3+\frac{x^2}{2}-2x]$
$-\frac{1}{2}[(3x^2+x-2)\frac{\sin (6x+2)}{6}+(6x+1)\frac{\cos (6x+2)}{36}-\frac{\sin (6x+2)}{6\times 36}]+C$
$-\frac{1}{2}[x^3+\frac{x^2}{2}-2x]-\frac{18x^2+6x-13}{72}\sin (6x+2)-\frac{1}{72}(6x+1)\cos (6x+2)+C.$