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Evaluate $\int \frac{5}{2}\mathrm{cos}\left(4x\right)\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)dx$.
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Emmy Sparks

Simplify.

$\int \frac{5\mathrm{cos}\left(4x\right)\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)}{2}dx$

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

$\frac{5}{2}\int \mathrm{cos}\left(4x\right)\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)dx$

Integrate by parts using the formula $\int udv=uv-\int vdu$, where $u=\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)$ and $dv=\mathrm{cos}\left(4x\right)$.

$\frac{5}{2}\left(\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\left(\frac{1}{4}\mathrm{sin}\left(4x\right)\right)-\int \frac{1}{4}\mathrm{sin}\left(4x\right)\frac{4\mathrm{cos}\left(4x\right)}{1+\mathrm{sin}\left(4x\right)}dx\right)$

Simplify.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\int \frac{\mathrm{sin}\left(4x\right)\mathrm{cos}\left(4x\right)}{1+\mathrm{sin}\left(4x\right)}dx\right)$

Let ${u}_{2}=\mathrm{sin}\left(4x\right)$. Then $d{u}_{2}=4\mathrm{cos}\left(4x\right)dx$$\frac{1}{4}d{u}_{2}=\mathrm{cos}\left(4x\right)dx$. Rewrite using ${u}_{2}$ and $d$${u}_{2}$.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\int \frac{{u}_{2}}{1+{u}_{2}}\cdot \frac{1}{4}d{u}_{2}\right)$

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\int \frac{{u}_{2}}{4\left(1+{u}_{2}\right)}d{u}_{2}\right)$

Since $\frac{1}{4}$ is constant with respect to ${u}_{2}$, move $\frac{1}{4}$ out of the integral.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\left(\frac{1}{4}\int \frac{{u}_{2}}{1+{u}_{2}}d{u}_{2}\right)\right)$

Reorder $1$ and ${u}_{2}$.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{1}{4}\int \frac{{u}_{2}}{{u}_{2}+1}d{u}_{2}\right)$

Divide ${u}_{2}$ by ${u}_{2}+1$.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{1}{4}\int 1-\frac{1}{{u}_{2}+1}d{u}_{2}\right)$

Split the single integral into multiple integrals.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{1}{4}\left(\int d{u}_{2}+\int -\frac{1}{{u}_{2}+1}d{u}_{2}\right)\right)$

Apply the constant rule.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{1}{4}\left({u}_{2}+C+\int -\frac{1}{{u}_{2}+1}d{u}_{2}\right)\right)$

Since $-1$ is constant with respect to ${u}_{2}$, move $-1$ out of the integral.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{1}{4}\left({u}_{2}+C-\int \frac{1}{{u}_{2}+1}d{u}_{2}\right)\right)$

Let ${u}_{3}={u}_{2}+1$. Then $d{u}_{3}=d{u}_{2}$. Rewrite using ${u}_{3}$ and $d$${u}_{3}$.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{1}{4}\left({u}_{2}+C-\int \frac{1}{{u}_{3}}d{u}_{3}\right)\right)$

The integral of 3$\frac{1}{{u}_{3}}$ with respect to ${u}_{3}$ is $\mathrm{ln}\left(|{u}_{3}|\right)$.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{1}{4}\left({u}_{2}+C-\left(\mathrm{ln}\left(|{u}_{3}|\right)+C\right)\right)\right)$

Simplify.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{{u}_{2}}{4}+\frac{\mathrm{ln}\left(|{u}_{3}|\right)}{4}\right)+C$

Substitute back in for each integration substitution variable.

$\frac{5}{2}\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)\mathrm{sin}\left(4x\right)}{4}-\frac{\mathrm{sin}\left(4x\right)}{4}+\frac{\mathrm{ln}\left(|\mathrm{sin}\left(4x\right)+1|\right)}{4}\right)+C$

Simplify.

$\frac{5\left(\mathrm{sin}\left(4x\right)\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)-\mathrm{sin}\left(4x\right)+\mathrm{ln}\left(|\mathrm{sin}\left(4x\right)+1|\right)\right)}{8}+C$

Reorder terms.

$\frac{5}{8}\left(\mathrm{sin}\left(4x\right)\mathrm{ln}\left(\mathrm{sin}\left(4x\right)+1\right)-\mathrm{sin}\left(4x\right)+\mathrm{ln}\left(|\mathrm{sin}\left(4x\right)+1|\right)\right)+C$