current objective evaluate indefinite integra...
current objective evaluate indefinite integrals involving trigonometric functions question evaluate the indefinite integral given below. ∫tan(x)(6cos(x) - 6sec(x))dx sorry, thats incorrect. try again? ∫tan(x)(6cos(x) - 6sec(x))dx =
Answer
# Explanation:
## Step1: Expand the integrand
\[
\begin{align*}
\int\tan(x)(6\cos(x)- 6\sec(x))dx&=\int(6\tan(x)\cos(x)-6\tan(x)\sec(x))dx\\
&=\int(6\frac{\sin(x)}{\cos(x)}\cos(x)-6\frac{\sin(x)}{\cos(x)}\frac{1}{\cos(x)})dx\\
&=\int(6\sin(x)-\frac{6\sin(x)}{\cos^{2}(x)})dx
\end{align*}
\]
## Step2: Split the integral
\[
\int(6\sin(x)-\frac{6\sin(x)}{\cos^{2}(x)})dx = 6\int\sin(x)dx-6\int\frac{\sin(x)}{\cos^{2}(x)}dx
\]
## Step3: Integrate term - by - term
For the first integral: $\int\sin(x)dx=-\cos(x)+C_1$.
For the second integral, let $u = \cos(x)$, then $du=-\sin(x)dx$. So $\int\frac{\sin(x)}{\cos^{2}(x)}dx=-\int\frac{du}{u^{2}}=\frac{1}{u}+C_2=\frac{1}{\cos(x)}+C_2$.
\[
\begin{align*}
6\int\sin(x)dx-6\int\frac{\sin(x)}{\cos^{2}(x)}dx&=6(-\cos(x))-6(\frac{1}{\cos(x)})+C\\
&=- 6\cos(x)-\frac{6}{\cos(x)}+C
\end{align*}
\]
# Answer:
$-6\cos(x)-\frac{6}{\cos(x)}+C$