2 Which of the following is a primitive of $\frac{\sin x}{\cos ^{3} x}$ ?
A. $\frac{1}{2} \sec ^{2} x$
B. $-\frac{1}{2} \sec ^{2} x$
C. $\frac{1}{4} \sec ^{4} x$
D. $-\frac{1}{4} \sec ^{4} x$

Answer

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Answer

So, the answer is $\boxed{\text{A}}$

Steps

Step 1 ：Let $u = \cos x$, then $\frac{du}{dx} = -\sin x$

Step 2 ：So, $\frac{\sin x}{\cos^3 x} dx = -\frac{1}{u^3} du$

Step 3 ：Now, we integrate $-\frac{1}{u^3} du$

Step 4 ：The integral is $\int -\frac{1}{u^3} du = \frac{1}{2u^2} + C$

Step 5 ：Substitute back $u = \cos x$, we get $\frac{1}{2\cos^2 x} + C$

Step 6 ：Finally, we have $\frac{1}{2\sec^2 x} + C$

Step 7 ：So, the answer is $\boxed{\text{A}}$