current objective evaluate indefinite integrals involving trigonometric functions question evaluate the…

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:

$-6\cos(x)-\frac{6}{\cos(x)}+C$