MHB Integrating a Product of Trig Functions

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The discussion centers on evaluating the integral of a product of trigonometric functions, specifically $$\int_{0}^{\pi/2}\d{}{x} \left(\sin\left({\frac{x}{2}}\right)\cos\left({\frac{x}{3}}\right)\right)\,dx$$. The provided answer from a TI calculator is $\frac{\sqrt{6}}{4}$. Participants highlight the use of the product rule for differentiation, noting that it complicates the problem. The fundamental theorem of calculus is referenced, emphasizing that differentiating and then integrating returns to the original function. The conversation concludes with a reiteration of the integral's evaluation process.
karush
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$$\int_{0}^{\pi/2}\d{}{x} \left(\sin\left({\frac{x}{2}}\right)\cos\left({\frac{x}{3}}\right)\right)\,dx$$

the ans the TI gave me was $\frac{\sqrt{6}}{4}$

the derivative can by found by the product rule. but really expands the problem
so not sure how the $\frac{d}{dx}$ played in this.
 
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karush said:
$$\int_{0}^{\pi/2}\d{}{x} \left(\sin\left({\frac{x}{2}}\right)\cos\left({\frac{x}{3}}\right)\right)\,dx$$

the ans the TI gave me was $\frac{\sqrt{6}}{4}$

the derivative can by found by the product rule. but really expands the problem
so not sure how the $\frac{d}{dx}$ played in this.
Fundamental theorem of calculus: if you differentiate and then integrate, you get back to where you started from! $$\int_{0}^{\pi/2}\d{}{x} \left(\sin\left({\frac{x}{2}}\right)\cos\left({\frac{x}{3}}\right)\right)\,dx = \left[\left(\sin\left({\frac{x}{2}}\right)\cos\left({\frac{x}{3}}\right)\right)\right]_0^{\pi/2}$$
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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