The given function is \( \displaystyle \sin^{3}x \cdot \cos^{3}x \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \( \int \sin^{3}x \cdot \cos^{3}x \, dx \).
Solution 1:
Let \(u = \sin x \), then \(du = \cos x \, dx \). Rewrite the integral:
\( \begin{aligned} & \int \sin^{3} x \cos^{3} x \, dx \\ &= \int \sin^{3} x \cos^{2} x \cdot \cos x \, dx \\ &= \int u^{3} (1 – u^{2}) \, du \quad \because \cos^{2} x = 1 – \sin^{2} x = 1 – u^{2} \\ &= \int (u^{3} – u^{5}) \, du \\ &= \frac{u^{4}}{4} – \frac{u^{6}}{6} + C, \quad \text{C is a constant} \\ &= \frac{\sin^{4} x}{4} – \frac{\sin^{6} x}{6} + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \sin^{3} x \cos^{3} x \, dx = \frac{\sin^{4} x}{4} – \frac{\sin^{6} x}{6} + C } \)Solution 2:
Let \(u = \cos x \), then \(du = -\sin x \, dx \). Rewrite the integral:
\( \begin{aligned} & \int \sin^{3} x \cos^{3} x \, dx \\ &= \int \sin^{2} x \cos^{3} x \cdot \sin x \, dx \\ &= -\int (1 – u^{2}) u^{3} \, du \quad \because \sin^{2} x = 1 – \cos^{2} x = 1 – u^{2} \\ &= -\int (u^{3} – u^{5}) \, du \\ &= -\left( \frac{u^{4}}{4} – \frac{u^{6}}{6} \right) + C, \quad \text{C is a constant} \\ &= \frac{u^{6}}{6} – \frac{u^{4}}{4} + C \\ &= \frac{\cos^{6} x}{6} – \frac{\cos^{4} x}{4} + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \sin^{3} x \cos^{3} x \, dx = \frac{\cos^{6} x}{6} – \frac{\cos^{4} x}{4} + C } \)Solution 3:
We know that \( \sin x \cos x = \frac{1}{2}\sin(2x)\) and \( \displaystyle \sin^3 x = \frac{3 \sin x \,- \sin(3x)}{4} \). Rewrite the integral:
\( \begin{aligned} & \int \sin^{3} x \cos^{3} x \, dx \\ &= \int (\sin x \cos x)^3 \, dx \\ &= \int \left( \frac{1}{2} \sin(2x) \right)^3 \, dx \\ &= \frac{1}{8} \int \sin^3(2x)\, dx \\ &= \frac{1}{8} \int \left( \frac{3 \sin(2x)\, – \sin(6x)}{4} \right) \, dx \\ &= \frac{1}{32} \int \left( 3 \sin(2x)\, – \sin(6x) \right) \, dx \\ &= \frac{1}{32} \left(3 \int \sin(2x) \, dx \, – \int \sin(6x) \, dx \right) \\ &= \frac{1}{32} \left( 3 \cdot \left( -\frac{1}{2} \cos(2x) \right) – \left( -\frac{1}{6} \cos(6x) \right) \right) + C, \quad \text{C is a constant} \\ &= \frac{1}{32} \left( -\frac{3}{2} \cos(2x) + \frac{1}{6} \cos(6x) \right) + C\\ &= = -\frac{3}{64} \cos(2x) + \frac{1}{192} \cos(6x) + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \sin^3 x \cos^3 x \, dx = -\frac{3}{64} \cos(2x) + \frac{1}{192} \cos(6x) + C} \)Please let me know in the comments if you find any errors in this solution.