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Simplify.

${\displaystyle \int \frac{7t{\cos }^{-1}\left(t\right)}{6}dt}$

Since ${\displaystyle \frac{7}{6}}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle \frac{7}{6}}$ out of the integral.

${\displaystyle \frac{7}{6}\int t{\cos }^{-1}\left(t\right)dt}$

Integrate by parts using the formula ${\displaystyle \int udv=uv-\int vdu}$, where ${\displaystyle u={\cos }^{-1}\left(t\right)}$ and ${\displaystyle dv=t}$.

${\displaystyle \frac{7}{6}({\cos }^{-1}\left(t\right)\left(\frac{1}{2}{t}^2\right)-\int \frac{1}{2}{t}^2(-\frac{1}{\sqrt{1-t^2}})dt)}$

Simplify.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}-\int -\frac{t^2}{2\sqrt{1-t^2}}dt)}$

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

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}--\int \frac{t^2}{2\sqrt{1-t^2}}dt)}$

Simplify.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\int \frac{t^2}{2\sqrt{1-t^2}}dt)}$

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

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{2}\int \frac{t^2}{\sqrt{1-t^2}}dt)}$

Let ${\displaystyle t=\sin \left(t_1\right)}$, where ${\displaystyle -\frac{\pi }{2}\le t_1\le \frac{\pi }{2}}$. Then ${\displaystyle dt=\cos \left(t_1\right)d{t}_1}$. Note that since ${\displaystyle -\frac{\pi }{2}\le t_1\le \frac{\pi }{2}}$, ${\displaystyle \cos \left(t_1\right)}$ is positive.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{2}\int \frac{{\sin }^2\left(t_1\right)}{\sqrt{1-{\sin }^2\left(t_1\right)}}\cos \left(t_1\right)d{t}_1)}$

Simplify terms.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{2}\int {\sin }^2\left(t_1\right)d{t}_1)}$

Use the half-angle formula to rewrite ${\displaystyle {\sin }^2\left(t_1\right)}$ as ${\displaystyle \frac{1-\cos \left(2t_1\right)}{2}}$.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{2}\int \frac{1-\cos \left(2t_1\right)}{2}d{t}_1)}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle t_1}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{2}(\frac{1}{2}\int 1-\cos \left(2t_1\right)d{t}_1))}$

Simplify.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{4}\int 1-\cos \left(2t_1\right)d{t}_1)}$

Split the single integral into multiple integrals.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{4}(\int d{t}_1+\int -\cos \left(2t_1\right)d{t}_1))}$

Apply the constant rule.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{4}(t_1+C+\int -\cos \left(2t_1\right)d{t}_1))}$

Since ${\displaystyle -1}$ is constant with respect to ${\displaystyle t_1}$, move ${\displaystyle -1}$ out of the integral.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{4}(t_1+C-\int \cos \left(2t_1\right)d{t}_1))}$

Let ${\displaystyle u=2t_1}$. Then ${\displaystyle du=2d{t}_1}$, so ${\displaystyle \frac{1}{2}du=d{t}_1}$. Rewrite using ${\displaystyle u}$ and ${\displaystyle d}$${\displaystyle u}$.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{4}(t_1+C-\int \cos \left(u\right)\frac{1}{2}du))}$

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

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{4}(t_1+C-\int \frac{\cos \left(u\right)}{2}du))}$

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

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{4}(t_1+C-(\frac{1}{2}\int \cos \left(u\right)du)))}$

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

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{1}{4}(t_1+C-\frac{1}{2}(\sin \left(u\right)+C)))}$

Simplify.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{t_1}{4}-\frac{\sin \left(u\right)}{8})+C}$

Substitute back in for each integration substitution variable.

${\displaystyle \frac{7}{6}(\frac{{\cos }^{-1}\left(t\right)t^2}{2}+\frac{{\sin }^{-1}\left(t\right)}{4}-\frac{\sin \left(2{\sin }^{-1}\left(t\right)\right)}{8})+C}$

Simplify.

$\frac{7}{6}(\frac{1}{2}{\cos }^{-1}\left(t\right)t^2+\frac{1}{4}{\sin }^{-1}\left(t\right)-\frac{1}{8}\sin \left(2{\sin }^{-1}\left(t\right)\right))+C$