Question

take the Intergral of (15t^-2 - 5t) cos(6t^-1 +t^2). I know I have to use sub rule. How do get from 6t^-1 +t^2) to -2(5/5)(3t+t)? It seems that if I multiplied this out i wont get 15t^-2 etc instead ill get 6t.

Answer 1

$$\begin{align*}\int\left(\frac{15}{t^2}-5t\right)\cos\left(\frac6t+t^2\right)\mathrm{d}t&=\frac15\int\left(\frac3{t^2}-t\right)\cos\left(\frac6t+t^2\right)\mathrm{d}t\\&=\frac25\int\left(\frac6{t^2}-2t\right)\cos\left(\frac6t+t^2\right)\mathrm{d}t\\&=-\frac25\int\left(-\frac6{t^2}+2t\right)\cos\left(\frac6t+t^2\right)\mathrm{d}t\end{align*}$$Let \(u=\frac6t+t^2\) so \(\mathrm{d}u=\left(-\frac6{t^2}+2t\right)\mathrm{d}t\).$$\begin{align*}-\frac25\int\left(-\frac6{t^2}+2t\right)\cos\left(\frac6t+t^2\right)\mathrm{d}t&=-\frac25\int\cos u\mathrm{d}u\\&=-\frac25\sin u+C\\&=-\frac25\sin\left(\frac6t+t^2\right)+C\end{align*}$$

Answer 2

@oldrin.bataku First two steps: should be a 5/2 in front, but the idea's basically the same.

Answer 3

@SithsAndGiggles oops!