Question

Integrate the following
1/1-sin x​

Answer 1

Answer:

$\\\tt tanx + sec x +C$

Explanation:

Been given:

$\\\tt \int \dfrac{1}{1-sinx} dx$

Multiply and divide by $\\\tt 1+sinx$

$\\\tt \int \dfrac{1}{1-sinx} dx \times \dfrac{1+sinx}{1+sinx} dx\\ \\\tt = \int \dfrac{1+sinx}{1-sin^2x} dx\\\\\tt = \int \dfrac{1+sinx}{cos^2x} dx \\\\\tt = \int sec^2 dx + \int secx \ tanx dx\\\\\tt = tanx + sec x +C$