Integral 7 6 &#xFEFF; </mrow> cos &#x2061;<!-- ⁡ 5 t <mo stret

Simplify.

${\displaystyle \int \frac{7\cos \left(5t\right)\ln (\sin \left(5t\right)+2)}{6}dt}$

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

${\displaystyle \frac{7}{6}\int \cos \left(5t\right)\ln (\sin \left(5t\right)+2)dt}$

Integrate by parts using the formula ${\displaystyle \int udv=uv-\int vdu}$, where ${\displaystyle u=\ln (\sin \left(5t\right)+2)}$ and ${\displaystyle dv=\cos \left(5t\right)}$.

${\displaystyle \frac{7}{6}(\ln (\sin \left(5t\right)+2)\left(\frac{1}{5}\sin \left(5t\right)\right)-\int \frac{1}{5}\sin \left(5t\right)\frac{5\cos \left(5t\right)}{2+\sin \left(5t\right)}dt)}$

Simplify.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\int \frac{\sin \left(5t\right)\cos \left(5t\right)}{2+\sin \left(5t\right)}dt)}$

Let ${\displaystyle u_2=\sin \left(5t\right)}$. Then ${\displaystyle d{u}_2=5\cos \left(5t\right)dt}$, so ${\displaystyle \frac{1}{5}d{u}_2=\cos \left(5t\right)dt}$. Rewrite using ${\displaystyle u_2}$ and ${\displaystyle d}$${\displaystyle u_2}$.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\int \frac{u_2}{2+u_2}\cdot \frac{1}{5}d{u}_2)}$

Simplify.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\int \frac{u_2}{5(2+u_2)}d{u}_2)}$

Since ${\displaystyle \frac{1}{5}}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle \frac{1}{5}}$ out of the integral.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-(\frac{1}{5}\int \frac{u_2}{2+u_2}d{u}_2))}$

Reorder ${\displaystyle 2}$ and ${\displaystyle u_2}$.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}\int \frac{u_2}{u_2+2}d{u}_2)}$

Divide ${\displaystyle u_2}$ by ${\displaystyle u_2+2}$.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}\int 1-\frac{2}{u_2+2}d{u}_2)}$

Split the single integral into multiple integrals.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}(\int d{u}_2+\int -\frac{2}{u_2+2}d{u}_2))}$

Apply the constant rule.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}(u_2+C+\int -\frac{2}{u_2+2}d{u}_2))}$

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

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}(u_2+C-\int \frac{2}{u_2+2}d{u}_2))}$

Since ${\displaystyle 2}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle 2}$ out of the integral.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}(u_2+C-(2\int \frac{1}{u_2+2}d{u}_2)))}$

Multiply ${\displaystyle 2}$ by ${\displaystyle -1}$.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}(u_2+C-2\int \frac{1}{u_2+2}d{u}_2))}$

Let ${\displaystyle u_3=u_2+2}$. Then ${\displaystyle d{u}_3=d{u}_2}$. Rewrite using ${\displaystyle u_3}$ and ${\displaystyle d}$${\displaystyle u_3}$.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}(u_2+C-2\int \frac{1}{u_3}d{u}_3))}$

The integral of ${\displaystyle \frac{1}{u_3}}$ with respect to ${\displaystyle u_3}$ is ${\displaystyle \ln \left(\left|u_3\right|\right)}$.${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{1}{5}(u_2+C-2(\ln \left(\left|u_3\right|\right)+C)))}$

Simplify.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{u_2}{5}+\frac{2\ln \left(\left|u_3\right|\right)}{5})+C}$

Substitute back in for each integration substitution variable.

${\displaystyle \frac{7}{6}(\frac{\ln (\sin \left(5t\right)+2)\sin \left(5t\right)}{5}-\frac{\sin \left(5t\right)}{5}+\frac{2\ln \left(|\sin \left(5t\right)+2|\right)}{5})+C}$

Simplify.

${\displaystyle \frac{7(\sin \left(5t\right)\ln (\sin \left(5t\right)+2)-\sin \left(5t\right)+2\ln \left(|\sin \left(5t\right)+2|\right))}{30}+C}$

Reorder terms.

$\frac{7}{30}(\sin \left(5t\right)\ln (\sin \left(5t\right)+2)-\sin \left(5t\right)+2\ln \left(|\sin \left(5t\right)+2|\right))+C$