Question

Integrating \(\int \cos ^{j} x \sin ^{k} x d x\) where \(k\) is Odd Evaluate \(\int \cos ^{8} x \sin ^{5} x d x\)

Short answer

The integral evaluates to \[-\left( \frac{\cos^9 x}{9} - 2\frac{\cos^{11} x}{11} + \frac{\cos^{13} x}{13} \right) + C.\]

Step-by-step solution

Identify Substitute Variables

For the integral \( \int \cos^{8} x \sin^{5} x \, dx \), since \( k = 5 \) is odd, we can make a substitution to simplify the integral. We let \( u = \cos x \), then \( du = -\sin x \, dx \).

Rearrange the Integral

Since one power of \( \sin x \) can be separated out as \( \sin x \, dx \) becomes \( -du \), rewrite the integral:\[\int \cos^{8} x \sin^{5} x \, dx = \int \cos^{8} x \sin^{4} x \cdot \sin x \, dx = -\int (\cos x)^{8}(1 - \cos^{2} x)^{2} \, du.\]

Simplify Using Trigonometric Identity

Simplify \( \sin^{4} x \) using the identity \( \sin^{2} x = 1 - \cos^{2} x \), so \( \sin^{4} x = (1 - \cos^{2} x)^{2} \). Substitute into the integral:\[-\int \cos^{8} x (1 - \cos^{2} x)^{2} \, du = -\int u^{8} (1-u^{2})^{2} \, du.\]

Expand the Polynomial

Expand the polynomial \((1-u^{2})^{2}\) to simplify the integral:\[(1-u^{2})^{2} = 1 - 2u^{2} + u^{4}.\]Thus, the integral becomes:\[-\int u^{8} (1 - 2u^{2} + u^{4}) \, du = -\int (u^{8} - 2u^{10} + u^{12}) \, du.\]

Integrate the Polynomial

Integrate each term separately:\[-\left( \int u^{8} \, du - 2\int u^{10} \, du + \int u^{12} \, du \right)\]Compute each integral:\[\int u^{8} \, du = \frac{u^{9}}{9}, \quad \int u^{10} \, du = \frac{u^{11}}{11}, \quad \int u^{12} \, du = \frac{u^{13}}{13}.\]Thus the expression becomes:\[-\left( \frac{u^{9}}{9} - 2\frac{u^{11}}{11} + \frac{u^{13}}{13} \right) + C.\]

Substitute Back to Original Variable

Substitute \( u = \cos x \) back into the integrated expression:\[-\left( \frac{\cos^9 x}{9} - 2\frac{\cos^{11} x}{11} + \frac{\cos^{13} x}{13} \right) + C.\]This is the solution to \( \int \cos^{8} x \sin^{5} x \, dx \).

Key concepts

Integration Techniques

Integrating functions that involve trigonometric expressions such as \( \int \cos^{8} x \sin^{5} x \, dx \) requires specific integration techniques. These techniques help break down complex functions into more manageable parts. One effective method is the use of substitution. In this case, since the power of \( \sin x \) is odd, it guides us to choose substitution for simplification. By substituting a variable, we transform the integral into a polynomial form that is easier to integrate. Another technique involved is rearranging the integral by factoring out a portion of the expression, like \( \sin x \cdot dx \), to match the derivative \( du \) of the chosen substitution variable. These methods collectively simplify the function, making integration feasible.

Trigonometric Identities

Trigonometric identities are essential tools for simplifying integrals involving trigonometric functions. In our example, the identity used is \( \sin^{2}x = 1 - \cos^{2}x \). This identity is crucial as it transforms the integral from involving both sine and cosine into one that is purely in terms of cosine. Specifically, when \( \sin^{5} x \) is odd and separated into \( \sin^{4} x \cdot \sin x \), we rewrite \( \sin^{4} x \) using the identity to become \( (1 - \cos^{2} x)^{2} \). These identities convert the problem into a polynomial composed of powers of cosine, facilitating further simplification and integration. They are invaluable for solving integrals where direct integration isn't possible.

Integration by Substitution

Integration by substitution is a fundamental technique for transforming a challenging integral into a simpler one by using derivatives and chain rules. In the given problem, due to \( k = 5 \) being odd, we choose \( u = \cos x \) and consequently, \( du = -\sin x \, dx \).
This technique essentially re-expresses the original function, swapping variables to reflect a simpler form. By substituting \( u \), we are able to express powers of both sine and cosine solely in terms of \( u \) which is then expanded and integrated as a polynomial. The core idea is to make a substitution that turns the integral into a form that is easily integrable. Later, the substitution is reversed to return the solution to the original variable, connecting back to the function we started with. This process highlights the power and utility of substitution in the realm of integration.