Question

The value of the integral \(\int\frac{1-\sin x}{\cos^2 x}dx\) is :

• A. secx - tanx + C, Where C is an arbitrary constant.
• B. tanx - secx + C, Where C is an arbitrary constant.
• C. secx tanx + C, Where C is an arbitrary constant.
• D. tanx + secx + C, Where C is an arbitrary constant.

Answer 1

The Correct Option is B

The given integral is: \(\int\frac{1-\sin x}{\cos^2 x}dx\).

To solve this, we first rewrite the integrand: \(\frac{1-\sin x}{\cos^2 x}=\frac{1}{\cos^2 x}-\frac{\sin x}{\cos^2 x}\).

This can be further expressed as: \(\sec^2 x-\sec x\tan x\).

Therefore, our integral becomes:

\(\int(\sec^2 x-\sec x\tan x)dx\)

Split this into two separate integrals:

\(\int\sec^2 xdx - \int\sec x\tan xdx\)

We know:

1. \(\int\sec^2 xdx=\tan x+C_1\)
2. \(\int\sec x\tan xdx=\sec x+C_2\)

Thus, combining these, we get:

\(\tan x-\sec x+C\), where \(C=C_1-C_2\) is a constant.

Hence, the value of the integral is tanx - secx + C, where C is an arbitrary constant.