Given the integral on the picture, we can use integration by parts to find the antiderivative.
First, let u = t² an dv = cos3t dt. If we find the derivative of u and the integral of v, we get:
![\begin{gathered} u=t²\Rightarrow du=2tdt \\ dv=cos3tdt\Rightarrow v=(1)/(3)sin3t \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/suv2ysai5l2fg3l0hxphpiv8tk5zayfwzz.png)
then, using the formula for integration by parts, we have the following:
![\int t²cos3tdt=(1)/(3)t²sin3t-\int(2)/(3)tsin3tdt](https://img.qammunity.org/2023/formulas/mathematics/college/oms7w2v689oa03w480yssnvy1sm2sa1fp0.png)
notice that the resulting integral on the right side also can be solved by parts. The solution of this integral is the following:
![\int(2)/(3)tsin3tdt=(2)/(9)tcos3t-(2)/(27)sin3t](https://img.qammunity.org/2023/formulas/mathematics/college/xf46hzmqi1fn2k4g6pmc5uliq5hr91wgf1.png)
then, combining both results, we get:
![\int t²cos3tdt=(1)/(3)t²sin3t+(2)/(9)tcos3t-(2)/(27)sin3t+C](https://img.qammunity.org/2023/formulas/mathematics/college/pela4zeydo5oarl5ezto8035ru98ilbmj3.png)