Question

$\int(\frac{1- sinx}{cos^2x})dx =$

• A. $$=tanx + secx +C$$
• B. $$=tanx - cosecx +C$$
• C. $$=tanx - secx +C$$
• D. $$=cotx - secx +C$$

Answer 1

Answer:

$$=tanx - secx +C$$

$\int(\frac{1- sinx}{cos^2x})dx = \int (\frac{1}{cos^2x})dx - \int(\frac{sinx}{cos^2x})dx$

$=\int (sec^2x)dx - \int \frac{sinx}{cosx} \times\frac{1}{cosx}dx = \int (sec^2x)dx - \int (secxtanx)$

$=tanx - secx +C$