Question

Assertion (A) If the anti-derivative of a function $\frac{2-3 \sin x}{\cos ^2 x}$ is $2 \sec x-3 \tan x+C$.
Reason (R) $\int \sec ^2 x d x=\tan x+C$ and $\int \tan x \sec x d x=\sec x+C$

• A. Both A and R are correct; R is the correct explanation of $A$
• B. Both A and R are correct; $R$ is not the correct explanation of $A$
• C. A is correct; $R$ is incorrect
• D. R is correct; A is incorrect

Answer 1

Answer: (D)

$ \int \frac{2-3 \sin x}{\cos ^2 x} d x $
$ \therefore \int \frac{2-3 \sin x}{\cos ^2 x} d x =\int \frac{2}{\cos ^2 x} d x-3 \int \frac{\sin x}{\cos ^2 x} d x $
$=2 \int \sec ^2 x d x-3 \int \tan x \cdot \sec x d x $
$ =2 \tan x-3 \sec x+C$