MHB Integrating a Product of Trig Functions

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! $$\int_{0}^{\pi/2}\d{}{x} \left(\sin\left({\frac{x}{2}}\right)\cos\left({\frac{x}{3}}\right)\right)\,dx = \left[\left(\sin\left({\frac{x}{2}}\right)\cos\left({\frac{x}{3}}\right)\right)\right]_0^{\pi/2}$$
 
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