`int_0^(pi/2)(sinx)^4/(sinx+cosx)dx`

Tuesday, March 6, 2012

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\left({\sin{{x}}}\right)}}^{{4}}}{{{\sin{{x}}}+{\cos{{x}}}}}{\left.{d}{x}\right.}$

The indefinite integral is complex. I don't want to work to the end, but there is an "industrial" way to convert this integral into an integral of a rational function: the tangent half-angle substitution.

Substitute $\displaystyle{t}={\tan{{\left(\frac{{x}}{{2}}\right)}}},$ then $\displaystyle{\sin{{\left({x}\right)}}}=\frac{{{2}{t}}}{{{1}+{{t}}^{{2}}}},$ $\displaystyle{\cos{{\left({x}\right)}}}=\frac{{{1}-{{t}}^{{2}}}}{{{1}+{{t}}^{{2}}}}$ and $\displaystyle{\left.{d}{x}\right.}=\frac{{{2}{\left.{d}{t}\right.}}}{{{1}+{{t}}^{{2}}}}.$ The limits $\displaystyle{\left[{0},\frac{\pi}{{2}}\right]}$ for $\displaystyle{x}$ become $\displaystyle{\left[{0},{1}\right]}$ for $\displaystyle{t},$ so the integral becomes

$\displaystyle{\int_{{0}}^{{1}}}{{\left(\frac{{{2}{t}}}{{{1}+{{t}}^{{2}}}}\right)}}^{{4}}\cdot\frac{{1}}{{\frac{{{1}+{2}{t}-{{t}}^{{2}}}}{{{1}+{{t}}^{{2}}}}}}\cdot\frac{{2}}{{{1}+{{t}}^{{2}}}}{\left.{d}{t}\right.}={32}{\int_{{0}}^{{1}}}\frac{{{{t}}^{{4}}{\left.{d}{t}\right.}}}{{{{\left({1}+{{t}}^{{2}}\right)}}^{{4}}{\left({1}+{2}{t}-{{t}}^{{2}}\right)}}},$

which may be with the general method for rational functions. The answer is about $\displaystyle{0.48}.$

Ninenotes

Tuesday, March 6, 2012

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{{{\left({\sin{{x}}}\right)}}^{{4}}}{{{\sin{{x}}}+{\cos{{x}}}}}{\left.{d}{x}\right.}$

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