Find the integral: − 1 <mrow> 2 t 2

Remove parentheses.

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

Split the single integral into multiple integrals.

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

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

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

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

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

Apply basic rules of exponents.

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

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

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

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

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

The integral of ${\displaystyle \frac{1}{t}}$ with respect to ${\displaystyle t}$ is ${\displaystyle \ln \left(\left|t\right|\right)}$.

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

Since ${\displaystyle \frac{3}{2}}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle \frac{3}{2}}$ out of the integral.

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

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

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

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

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

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

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

Simplify.

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

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

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

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

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

Since ${\displaystyle \frac{3}{2}}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle \frac{3}{2}}$ out of the integral.

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

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

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

Apply the constant rule.

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

Simplify.

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

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

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

Reorder terms.

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