What is $f (x) = \int x - x sec (\frac{x}{2}) d x$ $f(x) = int x-xsec(x/2) dx$ if $f (\frac{7 π}{4}) = 0$ $f((7pi)/4) = 0$ ?

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What is $f (x) = \int x - x sec (\frac{x}{2}) d x$ $f(x) = int x-xsec(x/2) dx$ if $f (\frac{7 π}{4}) = 0$ $f((7pi)/4) = 0$ ?

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Answer 1 · 2016-11-23T20:29:45

Because the indefinite integral involves the imaginary valued polylogarithmic function, I believe that there is an error in the question.

Explanation:

Wolfram Alpha gives the solution to the indefinite integral as

$4 i (L i_{2} (- i e^{\frac{i x}{2}}) - L i_{2} (i e^{\frac{i x}{2}})) + \frac{x^{2}}{2} - 2 x (ln (1 - i e^{\frac{i x}{2}}) - ln (1 + i e^{\frac{i x}{2}})) + C$ $4i(Li_2(-ie^((ix)/2)) - Li_2(ie^((ix)/2))) +x^2/2-2x(ln(1-ie^((ix)/2)) - ln(1+ie^((ix)/2))) +C$

Now substitute $\frac{7 π}{4}$ $(7pi)/4$ for $x$ $x$ , set equal to $0$ $0$ and solve for $C$ $C$ .

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What is $f (x) = \int x - x sec (\frac{x}{2}) d x$ $f(x) = int x-xsec(x/2) dx$ if $f (\frac{7 π}{4}) = 0$ $f((7pi)/4) = 0$ ?

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Explanation:

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What is f(x) = int x-xsec(x/2) dxf(x)=∫x−xsec(x2)dxf(x) = int x-xsec(x/2) dx if f((7pi)/4) = 0 f(7π4)=0f((7pi)/4) = 0 ?

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Explanation:

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What is $f (x) = \int x - x sec (\frac{x}{2}) d x$ $f(x) = int x-xsec(x/2) dx$ if $f (\frac{7 π}{4}) = 0$ $f((7pi)/4) = 0$ ?