Question

evaluate the integral using integration by parts.
int 8xsec^{2}2x dx
int 8xsec^{2}2x dx=square
(type an exact answer. use parentheses to clearly denote the argument

Explanation:

Step1: Recall integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = 8x$ and $\mathrm{d}v=\sec^{2}(2x)\mathrm{d}x$.

Step2: Find $\mathrm{d}u$ and $v$

Differentiate $u = 8x$ with respect to $x$ to get $\mathrm{d}u=8\mathrm{d}x$. Integrate $\mathrm{d}v=\sec^{2}(2x)\mathrm{d}x$ with respect to $x$. Let $t = 2x$, then $\mathrm{d}t = 2\mathrm{d}x$ and $\int\sec^{2}(2x)\mathrm{d}x=\frac{1}{2}\tan(2x)+C$, so $v=\frac{1}{2}\tan(2x)$.

Step3: Apply the integration - by - parts formula

$\int 8x\sec^{2}(2x)\mathrm{d}x=8x\times\frac{1}{2}\tan(2x)-\int\frac{1}{2}\tan(2x)\times8\mathrm{d}x=4x\tan(2x)-4\int\tan(2x)\mathrm{d}x$.

Step4: Integrate $\tan(2x)$

We know that $\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$. Let $u=\cos(2x)$, then $\mathrm{d}u=- 2\sin(2x)\mathrm{d}x$ and $\int\tan(2x)\mathrm{d}x=-\frac{1}{2}\ln|\cos(2x)|+C$.

Step5: Get the final result

$4x\tan(2x)-4\times(-\frac{1}{2}\ln|\cos(2x)|)+C=4x\tan(2x) + 2\ln|\cos(2x)|+C$.

Answer:

$4x\tan(2x)+2\ln|\cos(2x)|+C$