I am stuck on the last sentence. Please don't tell me what to do, just what they mean.
What equation do you get if you make the substitutions suggested? Can you solve the integrals on the RHS? Having solved them, you will have expressions involving sin(t) and cos(t) on both sides. Treat this as two equations, one involving only the sin terms and one involving only the cos terms.
Which sentence are you confused about?
Electronic calculators are not permitted? This just means you can't bring a calculator to use, you must use pencil and paper and the math tables they provide.
or are you talking about problem 13.
EDIT: Oops missed the Question 5 reference. Disregard my response...
I don't know what equation I'll get. How do I make the substitution?
"Use the expression (**) to find A and B by substituting for f(t) and f(x) in (*) and equating coefficients of sint and cost".
@Bolded, just what? how can i possibly 'substitute' f(t) and f(x) into (*)? if it means I set the expressions equal: [*] = [**] this doesn't yield anything.
I said it at the top of the OP: question 5 .
Basically, your second equation -- related to (**) -- into your first equation -- related to (*).
So, you need to solve for A and B in this equation:
Asin(t)+Bsin(t) = ∫(0->pi) [f(x) sin(x+t)]dx
Should be relatively simple. Just take the derivative of each side with respect to x, or you can parse out the integral of f(x)sin(x+t)... I would chose the first choice.
Edit: dont forget that A and B both have f(x) components if you try and take the derivative of the substituted equation.
I really don't see where this is going, both equations are identical:
sin(t).A+cos(t).B = ∫(0->pi) [f(x) sin(x+t)]dx
This is the same thing as:
sin(t).∫(0->pi)[f(x)cos(x)]dx +cos(t).∫(0->pi)[f(x)sinx]dx = ∫(0->pi) [f(x) sin(x+t)]dx
If I tried to differentiate I would just get:
sin(t)f(x)cos(x) dx + cos(t)f(x)sin(x) dx = f(x)sin(x+t) dx
You can't do anything with this equation.
I think you are approaching the derivative wrong.
please show me what you think the derivative should be.
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