I got stuck while trying to integrate this function, I've tried to make a new function so that I can solve the integral with a system of equations, but I couldn't get an answer. Here's the function $h(x)=\frac{8sin(x)+3cos(x)}{sin(x)+2cos(x)}$, and I am trying to calculate $\int_{0}^{\frac{\pi}{2}} h(x) dx$ I'd appreciate any help on this.
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try multiplying both the numerator by $\sec^3(x)$ – Jan 31 '19 at 21:36
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1https://math.stackexchange.com/questions/29980/evaluating-int-p-sin-x-cos-x-textdx – John Wayland Bales Jan 31 '19 at 21:38
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So, the idea is to split the integrand $\frac{8\sin x+3\cos x}{\sin x+2\cos x}$ into two parts, both of which will be easy to integrate. One of those, clearly, should be a multiple of $1=\frac{\sin x+2\cos x}{\sin x+2\cos x}$. The other? Well, it would be nice if we could just substitute $\sin x+2\cos x$. That works when the numerator is the derivative $\cos x - 2\sin x$, or a multiple of that.
So, there's our system: $$8\sin x+3\cos x=a(\sin x+2\cos x)+b(-2\sin x+\cos x)$$ Equate coefficients for $$a-2b = 8$$ $$2a+b = 3$$ Can you finish it from there?
jmerry
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