Integrate <mrow> 7 t </mrow> 2 &#xFEFF; </mrow> −

Split the single integral into multiple integrals.

${\displaystyle \int \frac{7t}{2}dt+\int -\frac{3}{t}dt+\int -2\cos \left(2t\right)dt+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

Since ${\displaystyle \frac{7}{2}}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle \frac{7}{2}}$ out of the integral.

${\displaystyle \frac{7}{2}\int tdt+\int -\frac{3}{t}dt+\int -2\cos \left(2t\right)dt+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

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

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)+\int -\frac{3}{t}dt+\int -2\cos \left(2t\right)dt+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

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

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-\int \frac{3}{t}dt+\int -2\cos \left(2t\right)dt+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

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

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-(3\int \frac{1}{t}dt)+\int -2\cos \left(2t\right)dt+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

Multiply ${\displaystyle 3}$ by ${\displaystyle -1}$.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3\int \frac{1}{t}dt+\int -2\cos \left(2t\right)dt+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

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

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)+\int -2\cos \left(2t\right)dt+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

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

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-2\int \cos \left(2t\right)dt+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

Let ${\displaystyle u_1=2t}$. Then ${\displaystyle d{u}_1=2dt}$, so ${\displaystyle \frac{1}{2}d{u}_1=dt}$. Rewrite using ${\displaystyle u_1}$ and ${\displaystyle d}$${\displaystyle u_1}$.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-2\int \cos \left(u_1\right)\frac{1}{2}d{u}_1+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

Combine ${\displaystyle \cos \left(u_1\right)}$ and ${\displaystyle \frac{1}{2}}$.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-2\int \frac{\cos \left(u_1\right)}{2}d{u}_1+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle u_1}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-2(\frac{1}{2}\int \cos \left(u_1\right)d{u}_1)+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

Simplify.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-\int \cos \left(u_1\right)d{u}_1+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

The integral of ${\displaystyle \cos \left(u_1\right)}$ with respect to ${\displaystyle u_1}$ is ${\displaystyle \sin \left(u_1\right)}$.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-(\sin \left(u_1\right)+C)+\int \frac{1}{2}\cdot \cos \left(3t\right)dt}$

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

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-(\sin \left(u_1\right)+C)+\frac{1}{2}\int \cos \left(3t\right)dt}$

Let ${\displaystyle u_2=3t}$. Then ${\displaystyle d{u}_2=3dt}$, so ${\displaystyle \frac{1}{3}d{u}_2=dt}$. Rewrite using ${\displaystyle u_2}$ and ${\displaystyle d}$${\displaystyle u_2}$.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-(\sin \left(u_1\right)+C)+\frac{1}{2}\int \cos \left(u_2\right)\frac{1}{3}d{u}_2}$

Combine ${\displaystyle \cos \left(u_2\right)}$ and ${\displaystyle \frac{1}{3}}$.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-(\sin \left(u_1\right)+C)+\frac{1}{2}\int \frac{\cos \left(u_2\right)}{3}d{u}_2}$

Since ${\displaystyle \frac{1}{3}}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle \frac{1}{3}}$ out of the integral.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-(\sin \left(u_1\right)+C)+\frac{1}{2}(\frac{1}{3}\int \cos \left(u_2\right)d{u}_2)}$

Simplify.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-(\sin \left(u_1\right)+C)+\frac{1}{6}\int \cos \left(u_2\right)d{u}_2}$

The integral of ${\displaystyle \cos \left(u_2\right)}$ with respect to ${\displaystyle u_2}$ is ${\displaystyle \sin \left(u_2\right)}$.

${\displaystyle \frac{7}{2}(\frac{1}{2}{t}^2+C)-3(\ln \left(\left|t\right|\right)+C)-(\sin \left(u_1\right)+C)+\frac{1}{6}(\sin \left(u_2\right)+C)}$

Simplify.

${\displaystyle \frac{7t^2}{4}-3\ln \left(\left|t\right|\right)-\sin \left(u_1\right)+\frac{1}{6}\sin \left(u_2\right)+C}$

Substitute back in for each integration substitution variable.

${\displaystyle \frac{7t^2}{4}-3\ln \left(\left|t\right|\right)-\sin \left(2t\right)+\frac{1}{6}\sin \left(3t\right)+C}$

Reorder terms.

$\frac{7}{4}{t}^2-3\ln \left(\left|t\right|\right)-\sin \left(2t\right)+\frac{1}{6}\sin \left(3t\right)+C$