Question

$\int \sin^3 x.\cos^6 x \, dx =$

• A. $\frac{\cos^7 x}{7} - \frac{\cos^9 x}{9} + c$
• B. $\frac{\sin^9 x}{9} - \frac{\sin^7 x}{7} + c$
• C. $\frac{\sin^7 x}{7} - \frac{\sin^9 x}{9} + c$
• D. $\frac{\cos^9 x}{9} - \frac{\cos^7 x}{7} + c$

Answer 1

Answer: (D)

$\frac{\cos^9 x}{9} - \frac{\cos^7 x}{7} + c$

Answer 2

The integral to be solved is $\int \sin^3 x.\cos^6 x \, dx$.

We can rewrite the integrand as: $\int \sin^2 x \cdot \cos^6 x \cdot \sin x \, dx$

Using the identity $\sin^2 x = 1 - \cos^2 x$, we substitute it into the integral: $\int (1 - \cos^2 x) \cdot \cos^6 x \cdot \sin x \, dx$

Now, let's use the substitution method. Let $u = \cos x$. Then, differentiate both sides with respect to $x$: $du = -\sin x \, dx$ This implies $\sin x \, dx = -du$.

Substitute $u = \cos x$ and $\sin x \, dx = -du$ into the integral: $\int (1 - u^2) \cdot u^6 \cdot (-du)$ $= -\int (u^6 - u^2 \cdot u^6) \, du$ $= -\int (u^6 - u^8) \, du$

Distribute the negative sign: $= \int (u^8 - u^6) \, du$

Now, integrate term by term using the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$: $= \frac{u^{8+1}}{8+1} - \frac{u^{6+1}}{6+1} + C$ $= \frac{u^9}{9} - \frac{u^7}{7} + C$

Finally, substitute back $u = \cos x$: $= \frac{\cos^9 x}{9} - \frac{\cos^7 x}{7} + C$

Comparing this result with the given options, it matches the fourth option.