Question

\(\int \frac{cos^3x}{sin^2x+sinx}\,dx =\)

• A. \(log(cosx)+cosx+c\), where c is the constant of integration
• B. \(log(sinx)-sinx+c\), where c is the constant of integration
• C. \(log(sinx)+sinx+c\), where c is the constant of integration
• D. \(log(cosx)-cosx+c\), where c is the constant of integration

Hint:

Substitute t = sin x and cancel the common factor 1 + t.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Write \(\cos^3x\,dx = \cos^2x\cdot\cos x\,dx = (1 - \sin^2x)\,d(\sin x)\).

Step 2: Substitute:
Let \(t = \sin x\), \(dt = \cos x\,dx\).
\[ \int\frac{1 - t^2}{t^2 + t}\,dt = \int\frac{(1-t)(1+t)}{t(t+1)}\,dt = \int\frac{1 - t}{t}\,dt \]

Step 3: Integrate:
\[ \int\left(\frac1t - 1\right)dt = \log|t| - t + c = \log(\sin x) - \sin x + c \]

Step 4: Check the options:
Option (B) matches. Options (A) and (D) use \(\cos x\) as the variable, and option (C) has the wrong sign on \(\sin x\).

Final Answer:
The integral is log(sin x) - sin x + c. \[ \boxed{\text{(B) }\log(\sin x)-\sin x+c} \]