Definite Integration 78C00D

1. **State the problem:** We need to find the indefinite integral $$\int x \cos x \, dx$$ and then evaluate the definite integral $$y = \int x \cos x \, dx$$ for some limits (not specified). 2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = x \implies du = dx$$ $$dv = \cos x \, dx \implies v = \sin x$$ 4. **Apply integration by parts:** $$\int x \cos x \, dx = x \sin x - \int \sin x \, dx$$ 5. **Integrate $$\int \sin x \, dx$$:** $$\int \sin x \, dx = -\cos x + C$$ 6. **Substitute back:** $$\int x \cos x \, dx = x \sin x - (-\cos x) + C = x \sin x + \cos x + C$$ 7. **Identify constants:** The indefinite integral can be written as: $$\int x \cos x \, dx = A \sin x + B$$ But from the result, it is: $$x \sin x + \cos x + C$$ So the given form $$A \sin x + B$$ does not match the integral of $$x \cos x$$. Possibly the problem meant $$\int x \sin x \, dx$$ for the indefinite integral. 8. **Check the indefinite integral $$\int x \sin x \, dx$$:** Using integration by parts: $$u = x, dv = \sin x \, dx \implies du = dx, v = -\cos x$$ Then: $$\int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C$$ This matches the form $$A \sin x + B$$ if we write: $$-x \cos x + \sin x + C = A \sin x + B$$ But since $$x$$ is present, it cannot be simplified to just $$A \sin x + B$$. 9. **Conclusion:** - For the indefinite integral $$\int x \cos x \, dx$$, the answer is: $$x \sin x + \cos x + C$$ - For the definite integral $$y = \int x \cos x \, dx$$ over limits $$a$$ to $$b$$: $$y = \left[ x \sin x + \cos x \right]_a^b = (b \sin b + \cos b) - (a \sin a + \cos a)$$ Since limits are not given, the definite integral value $$C$$ cannot be computed. **Final answers:** $$A = \text{not applicable}$$ $$B = \text{not applicable}$$ $$C = \text{depends on limits}$$