Question

evaluate the integral using integration by parts.
int 10\thetasec^{2}\theta d\theta
int 10\thetasec^{2}\theta d\theta=square
(use parentheses to clearly denote the argument of each function.)

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 $u = 10\theta$ and $\mathrm{d}v=\sec^{2}\theta\mathrm{d}\theta$.

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

Differentiate $u = 10\theta$ with respect to $\theta$ to get $\mathrm{d}u = 10\mathrm{d}\theta$. Integrate $\mathrm{d}v=\sec^{2}\theta\mathrm{d}\theta$ with respect to $\theta$ to get $v=\tan\theta$.

Step3: Apply the integration - by - parts formula

$\int 10\theta\sec^{2}\theta\mathrm{d}\theta=10\theta\tan\theta-\int 10\tan\theta\mathrm{d}\theta$.

Step4: Integrate $\int 10\tan\theta\mathrm{d}\theta$

We know that $\int\tan\theta\mathrm{d}\theta=-\ln|\cos\theta| + C$, so $\int 10\tan\theta\mathrm{d}\theta=- 10\ln|\cos\theta|+C$.

Answer:

$10\theta\tan\theta + 10\ln|\cos\theta|+C$