12. find the antiderivative $intsin(5x - 4)dx$ 13. evaluate the integral $int_{-1}^{0}\frac{t}{(2 +…

Explanation:

Step1: Use substitution for $\int\sin(5x - 4)dx$

Let $u = 5x-4$, then $du=5dx$, $dx=\frac{1}{5}du$. So $\int\sin(5x - 4)dx=\frac{1}{5}\int\sin(u)du$.

Step2: Integrate $\sin(u)$

We know that $\int\sin(u)du=-\cos(u)+C$. So $\frac{1}{5}\int\sin(u)du=-\frac{1}{5}\cos(u)+C$. Substituting back $u = 5x - 4$, we get $-\frac{1}{5}\cos(5x - 4)+C$.

Step3: Use substitution for $\int_{-1}^{0}\frac{t}{(2 + t^{2})^{4}}dt$

Let $v=2 + t^{2}$, then $dv = 2tdt$, $tdt=\frac{1}{2}dv$. When $t=-1$, $v=2 + 1=3$; when $t = 0$, $v=2+0 = 2$. So $\int_{-1}^{0}\frac{t}{(2 + t^{2})^{4}}dt=\frac{1}{2}\int_{3}^{2}v^{-4}dv$.

Step4: Integrate $v^{-4}$

$\int v^{-4}dv=\frac{v^{-4 + 1}}{-4+1}=-\frac{1}{3}v^{-3}+C$. Then $\frac{1}{2}\int_{3}^{2}v^{-4}dv=\frac{1}{2}\left[-\frac{1}{3}v^{-3}\right]_{3}^{2}$.

Step5: Evaluate the definite - integral

$\frac{1}{2}\left(-\frac{1}{3}v^{-3}\right)\big|_{3}^{2}=\frac{1}{2}\left(-\frac{1}{3\times2^{3}}+\frac{1}{3\times3^{3}}\right)=\frac{1}{2}\left(-\frac{1}{24}+\frac{1}{81}\right)=\frac{1}{2}\times\frac{-27 + 8}{648}=\frac{1}{2}\times\frac{-19}{648}=-\frac{19}{1296}$.

Answer:

1. The antiderivative of $\int\sin(5x - 4)dx$ is $-\frac{1}{5}\cos(5x - 4)+C$.
2. The value of $\int_{-1}^{0}\frac{t}{(2 + t^{2})^{4}}dt$ is $-\frac{19}{1296}$.