Find the integral: <mrow> 5 t 3 </mrow

Remove parentheses.

${\displaystyle \int \frac{5t^3}{2}+\frac{3}{t^3}-\frac{\sin \left(t\right)}{2}-2\cos \left(2t\right)+2dt}$

Split the single integral into multiple integrals.

${\displaystyle \int \frac{5t^3}{2}dt+\int \frac{3}{t^3}dt+\int -\frac{\sin \left(t\right)}{2}dt+\int -2\cos \left(2t\right)dt+\int 2dt}$

Since ${\displaystyle \frac{5}{2}}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle \frac{5}{2}}$ out of the integral.

${\displaystyle \frac{5}{2}\int t^{3}dt+\int \frac{3}{t^3}dt+\int -\frac{\sin \left(t\right)}{2}dt+\int -2\cos \left(2t\right)dt+\int 2dt}$

By the Power Rule, the integral of ${\displaystyle t^3}$ with respect to ${\displaystyle t}$ is ${\displaystyle \frac{1}{4}{t}^4}$.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+\int \frac{3}{t^3}dt+\int -\frac{\sin \left(t\right)}{2}dt+\int -2\cos \left(2t\right)dt+\int 2dt}$

Since ${\displaystyle 3}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle 3}$ out of the integral.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3\int \frac{1}{t^3}dt+\int -\frac{\sin \left(t\right)}{2}dt+\int -2\cos \left(2t\right)dt+\int 2dt}$

Apply basic rules of exponents.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3\int t^{-3}dt+\int -\frac{\sin \left(t\right)}{2}dt+\int -2\cos \left(2t\right)dt+\int 2dt}$

By the Power Rule, the integral of ${\displaystyle t^{-3}}$ with respect to ${\displaystyle t}$ is ${\displaystyle -\frac{1}{2}{t}^{-2}}$.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2}{t}^{-2}+C)+\int -\frac{\sin \left(t\right)}{2}dt+\int -2\cos \left(2t\right)dt+\int 2dt}$

Simplify.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)+\int -\frac{\sin \left(t\right)}{2}dt+\int -2\cos \left(2t\right)dt+\int 2dt}$

Since ${\displaystyle -1}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle -1}$ out of the integral.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\int \frac{\sin \left(t\right)}{2}dt+\int -2\cos \left(2t\right)dt+\int 2dt}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-(\frac{1}{2}\int \sin \left(t\right)dt)+\int -2\cos \left(2t\right)dt+\int 2dt}$

The integral of ${\displaystyle \sin \left(t\right)}$ with respect to ${\displaystyle t}$ is ${\displaystyle -\cos \left(t\right)}$.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\frac{1}{2}(-\cos \left(t\right)+C)+\int -2\cos \left(2t\right)dt+\int 2dt}$

Since ${\displaystyle -2}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle -2}$ out of the integral.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\frac{1}{2}(-\cos \left(t\right)+C)-2\int \cos \left(2t\right)dt+\int 2dt}$

Let ${\displaystyle u=2t}$. Then ${\displaystyle du=2dt}$, so ${\displaystyle \frac{1}{2}du=dt}$. Rewrite using ${\displaystyle u}$ and ${\displaystyle d}$${\displaystyle u}$.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\frac{1}{2}(-\cos \left(t\right)+C)-2\int \cos \left(u\right)\frac{1}{2}du+\int 2dt}$

Combine ${\displaystyle \cos \left(u\right)}$ and ${\displaystyle \frac{1}{2}}$.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\frac{1}{2}(-\cos \left(t\right)+C)-2\int \frac{\cos \left(u\right)}{2}du+\int 2dt}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle u}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\frac{1}{2}(-\cos \left(t\right)+C)-2(\frac{1}{2}\int \cos \left(u\right)du)+\int 2dt}$

Simplify.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\frac{1}{2}(-\cos \left(t\right)+C)-\int \cos \left(u\right)du+\int 2dt}$

The integral of ${\displaystyle \cos \left(u\right)}$ with respect to ${\displaystyle u}$ is ${\displaystyle \sin \left(u\right)}$.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\frac{1}{2}(-\cos \left(t\right)+C)-(\sin \left(u\right)+C)+\int 2dt}$

Apply the constant rule.

${\displaystyle \frac{5}{2}(\frac{1}{4}{t}^4+C)+3(-\frac{1}{2t^2}+C)-\frac{1}{2}(-\cos \left(t\right)+C)-(\sin \left(u\right)+C)+2t+C}$

Simplify.

${\displaystyle \frac{5t^4}{8}-\frac{3}{2t^2}+\frac{\cos \left(t\right)}{2}-\sin \left(u\right)+2t+C}$

Replace all occurrences of ${\displaystyle u}$ with ${\displaystyle 2t}$.

${\displaystyle \frac{5t^4}{8}-\frac{3}{2t^2}+\frac{\cos \left(t\right)}{2}-\sin \left(2t\right)+2t+C}$

Reorder terms.

$\frac{5}{8}{t}^4-\frac{3}{2t^2}+\frac{1}{2}\cos \left(t\right)-\sin \left(2t\right)+2t+C$