If f(x) = int csc^5 x dx,then f(frac{pi}{4}) | vedclass

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(C) We have $f(x) = \int \csc^5 x \ dx$. Using the reduction formula for $\int \csc^n x \ dx$,we know that $\int \csc^n x \ dx = -\frac{\csc^{n-2} x \

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$-\frac{1}{8} [7\sqrt{2} + 3\log(\sqrt{2} + 1)] + c$

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(C) We have $f(x) = \int \csc^5 x \ dx$. Using the reduction formula for $\int \csc^n x \ dx$,we know that $\int \csc^n x \ dx = -\frac{\csc^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1} \int \csc^{n-2} x \ dx$. For $n=5$,we get $f(x) = -\frac{\csc^3 x \cot x}{4} + \frac{3}{4} \int \csc^3 x \ dx$. We know $\int \csc^3 x \ dx = -\frac{1}{2} \csc x \cot x + \frac{1}{2} \log |\csc x - \cot x| + C$. Substituting this back,$f(x) = -\frac{1}{4} \csc^3 x \cot x + \frac{3}{4} [-\frac{1}{2} \csc x \cot x + \frac{1}{2} \log |\csc x - \cot x|] + c$. $f(x) = -\frac{1}{4} \csc^3 x \cot x - \frac{3}{8} \csc x \cot x + \frac{3}{8} \log |\csc x - \cot x| + c$. At $x = \frac{\pi}{4}$,$\csc(\frac{\pi}{4}) = \sqrt{2}$ and $\cot(\frac{\pi}{4}) = 1$. $f(\frac{\pi}{4}) = -\frac{1}{4} (\sqrt{2})^3 (1) - \frac{3}{8} (\sqrt{2})(1) + \frac{3}{8} \log |\sqrt{2} - 1| + c$. $f(\frac{\pi}{4}) = -\frac{2\sqrt{2}}{4} - \frac{3\sqrt{2}}{8} + \frac{3}{8} \log |\sqrt{2} - 1| + c$. $f(\frac{\pi}{4}) = -\frac{4\sqrt{2}}{8} - \frac{3\sqrt{2}}{8} + \frac{3}{8} \log |\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}| + c$. $f(\frac{\pi}{4}) = -\frac{7\sqrt{2}}{8} + \frac{3}{8} \log |\frac{1}{\sqrt{2}+1}| + c = -\frac{7\sqrt{2}}{8} - \frac{3}{8} \log(\sqrt{2}+1) + c$. $f(\frac{\pi}{4}) = -\frac{1}{8} [7\sqrt{2} + 3 \log(\sqrt{2} + 1)] + c$.

If $f(x) = \int \csc^5 x \ dx$,then $f(\frac{\pi}{4}) = $

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