Question

$\int \sin^5 x \,dx =$

• A. $\cos x + \frac{2}{3}\cos^2 x - \frac{\cos^5 x}{5} + c$, where c is the constant of integration
• B. $\cos x + \frac{2}{3}\cos^2 x + \frac{\cos^5 x}{5} + c$, where c is the constant of integration
• C. $-\left(\cos x - \frac{2}{3}\cos^3 x + \frac{\cos^5 x}{5}\right) + c$, where c is the constant of integration
• D. $\cos x - \frac{2}{3}\cos^2 x + \frac{\cos^5 x}{5} + c$, where c is the constant of integration

Hint:

For odd powers of sin/cos, peel off one to join the $dx$ and convert the rest to the other function.

Answer 1

The Correct Option is C

Step 1: Concept
For odd powers of $\sin x$, use the substitution $u = \cos x$.

Step 2: Meaning
$du = -\sin x dx$. Transform $\sin^5 x$ using $\sin^2 x = 1 - \cos^2 x$.

Step 3: Analysis
$\int \sin^4 x \sin x dx = \int (1 - \cos^2 x)^2 \sin x dx$. Let $u = \cos x \implies -\int (1 - u^2)^2 du = -\int (1 - 2u^2 + u^4) du$. $= -(u - \frac{2}{3}u^3 + \frac{1}{5}u^5) + c$. $= -(\cos x - \frac{2}{3}\cos^3 x + \frac{1}{5}\cos^5 x) + c$.

Step 4: Conclusion
Matches option (C). Final Answer: (C)