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How to evaluate these integrals: $\int_{-1}^{1}(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin5xdx$ $\int_{-1}^{1}\frac{(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin5x}{\sqrt{1-x^{2}}}dx$ Can anybody help me ? Thank you!

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    $\int_0^1x^n(1+x^2)^\frac{1}{4}\sin5x~dx=\int_0^{\sinh^{-1}1}\sinh^nx(1+\sinh^2x)^\frac{1}{4}\sin(5\sinh x)~d(\sinh x)=\int_0^{\sinh^{-1}1}(1+\sinh^2x)^\frac{1}{4}\sin(5\sinh x)\sinh^nx\cosh^\frac{3}{2}x~dx~,~\int_0^1\dfrac{x^n(1+x^2)^\frac{1}{4}\sin5x}{\sqrt{1-x^2}}dx=\int_0^{\sinh^{-1}1}\dfrac{\sinh^nx(1+\sinh^2x)^\frac{1}{4}\sin(5\sinh x)}{\sqrt{1-\sinh^2x}}d(\sinh x)=\int_0^{\sinh^{-1}1}\dfrac{\sin(5\sinh x)}{\sqrt{1-\sinh^2x}\sinh^nx\cosh^\frac{3}{2}x}dx$ think about incomplete types of bessel functions2014-10-28

2 Answers 2

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To refresh the definition here $f(x)$ is an odd function if $f(-x) = -f(x)$, and is even if $f(-x)=f(x)$

Firstly, $\int_{-1}^1 f(x) \hspace{4pt} dx = 0$ if $f(x)$ is odd function, and if $f(x)$ is an even function then

$\int_{-1}^1 f(x) dx = 2 \int_0^1 f(x) dx $

For the first problem:

$f(x) = (x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin5x$

$ \begin{align*} f(-x) &= (-x^{5} + x^{3} +x)(1+x^{2})^{\frac{1}{4}} \sin5(-x) \\ &= (-x^{5} + x^{3} +x)(1+x^{2})^{\frac{1}{4}} (-\sin 5x)\\ &= (x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin 5x = f(x)\\ \end{align*} $

For the first function if you use the approximation (use high order terms if you want ) $(1+x^2)^{\frac{1}{4}} \approx 1+\frac{1}{4}x^{2}$

The integral is

$ 2 \int_0^1 (x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}\sin5x dx \approx 2 \int_0^1 (x^{5}-x^{3}-x)(1+\frac{x^2}{4})\sin5x dx$

which is

$ 2 \int_0^1 \left( \frac{3}{4} x^5 \sin(5x) -x \sin(5x) - \frac{5x^3}{4} \sin(5x) + \frac{x^7}{4} \sin(5x) \right) dx $

$ \approx 2 \left( \frac{98165 \cos(5) -22233 \sin(5)}{312500} \right) = 0.314658 $

Where this can be verified on Wolfram Alpha

Similarly for the second integral

$ \int_{-1}^1 \frac{(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}}{\sqrt{1-x^2}}\sin5x dx = 2 \int_0^1 \frac{(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}}{\sqrt{1-x^2}}\sin5x dx$

Using another approximation

$ \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{\frac{-1}{2}} \approx (1+\frac{x^2}{2}) $

Therefore

$ \int_{-1}^1 \frac{(x^{5}-x^{3}-x)(1+x^{2})^{\frac{1}{4}}}{\sqrt{1-x^2}}\hspace{4pt}sin5x \hspace{4pt} dx = 2 \int_0^1 (x^{5}-x^{3}-x)(1+\frac{1}{4}x^{2})(1+\frac{1}{2}x^2)\sin5x dx$

$ = 2 \int_0^1 \left(\frac{x^9}{8} + \frac{x^7}{4} - \frac{x^5}{8} - \frac{7x^3}{8} -x \right)\sin5x dx $

$ = \frac{2}{15625000} \left( (703125x^8 - 481250 x^6 + 186875 x^4 -1730325 x^2 - 486574)\sin5x - 5x(78125 x^8 - 68750x^6 + 37375 x^4-576775x^2 -486574)\cos5x \right) $

with limits from $0$ to $1$

The value of second integral is therefore $0.406494$

(Note that the simplified function in the second integral is also an even function, and therefore the integral limits are from $0$ to $1$)

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For $\int_{-1}^1(x^5-x^3-x)(1+x^2)^\frac{1}{4}\sin5x~dx$ ,

$\int_{-1}^1(x^5-x^3-x)(1+x^2)^\frac{1}{4}\sin5x~dx$

$=2\int_0^1(x^5-x^3-x)(1+x^2)^\frac{1}{4}\sin5x~dx$

$=2\int_0^1x^5(1+x^2)^\frac{1}{4}\sin5x~dx-2\int_0^1x(1+x^2)^\frac{5}{4}\sin5x~dx$

$=\int_0^1x^4(1+x^2)^\frac{1}{4}\sin5x~d(x^2)-\int_0^1(1+x^2)^\frac{5}{4}\sin5x~d(x^2)$

$=\int_0^1(1+2x^2+x^4)(1+x^2)^\frac{1}{4}\sin5x~d(x^2)-\int_0^1(2+2x^2)(1+x^2)^\frac{1}{4}\sin5x~d(x^2)+\int_0^1(1+x^2)^\frac{1}{4}\sin5x~d(x^2)-\int_0^1(1+x^2)^\frac{5}{4}\sin5x~d(x^2)$

$=\int_0^1(1+x^2)^\frac{9}{4}\sin5x~d(x^2)-3\int_0^1(1+x^2)^\frac{5}{4}\sin5x~d(x^2)+\int_0^1(1+x^2)^\frac{1}{4}\sin5x~d(x^2)$

$=\dfrac{4}{13}\int_0^1\sin5x~d\left((1+x^2)^\frac{13}{4}\right)-\dfrac{4}{3}\int_0^1\sin5x~d\left((1+x^2)^\frac{9}{4}\right)+\dfrac{4}{5}\int_0^1\sin5x~d\left((1+x^2)^\frac{5}{4}\right)$

$=\dfrac{4}{13}\left[(1+x^2)^\frac{13}{4}\sin5x\right]_0^1-\dfrac{4}{13}\int_0^1(1+x^2)^\frac{13}{4}~d(\sin5x)-\dfrac{4}{3}\left[(1+x^2)^\frac{9}{4}\sin5x\right]_0^1+\dfrac{4}{3}\int_0^1(1+x^2)^\frac{9}{4}~d(\sin5x)+\dfrac{4}{5}\left[(1+x^2)^\frac{5}{4}\sin5x\right]_0^1-\dfrac{4}{5}\int_0^1(1+x^2)^\frac{5}{4}~d(\sin5x)$

$=\dfrac{2^\frac{21}{4}\sin5}{13}-\dfrac{2^\frac{17}{4}\sin5}{3}+\dfrac{2^\frac{13}{4}\sin5}{5}-\dfrac{20}{13}\int_0^1(1+x^2)^\frac{13}{4}\cos5x~dx+\dfrac{20}{3}\int_0^1(1+x^2)^\frac{9}{4}\cos5x~dx-4\int_0^1(1+x^2)^\frac{5}{4}\cos5x~dx$

$=\dfrac{2^\frac{21}{4}\sin5}{13}-\dfrac{2^\frac{17}{4}\sin5}{3}+\dfrac{2^\frac{13}{4}\sin5}{5}-\dfrac{20}{13}\int_0^{\sinh^{-1}1}(1+\sinh^2x)^\frac{13}{4}\cos(5\sinh x)~d(\sinh x)+\dfrac{20}{3}\int_0^{\sinh^{-1}1}(1+\sinh^2x)^\frac{9}{4}\cos(5\sinh x)~d(\sinh x)-4\int_0^{\sinh^{-1}1}(1+\sinh^2x)^\frac{5}{4}\cos(5\sinh x)~d(\sinh x)$

$=\dfrac{2^\frac{21}{4}\sin5}{13}-\dfrac{2^\frac{17}{4}\sin5}{3}+\dfrac{2^\frac{13}{4}\sin5}{5}-\dfrac{20}{13}\int_0^{\ln(1+\sqrt2)}\cos(5\sinh x)\cosh^\frac{15}{2}x~dx+\dfrac{20}{3}\int_0^{\ln(1+\sqrt2)}\cos(5\sinh x)\cosh^\frac{11}{2}x~dx-4\int_0^{\ln(1+\sqrt2)}\cos(5\sinh x)\cosh^\frac{7}{2}x~dx$

Which relates to incomplete Bessel function.

For $\int_{-1}^1\dfrac{(x^5-x^3-x)(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~dx$ ,

$\int_{-1}^1\dfrac{(x^5-x^3-x)(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~dx$

$=2\int_0^1\dfrac{(x^5-x^3-x)(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~dx$

$=2\int_0^1\dfrac{x^5(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~dx-2\int_0^1\dfrac{x(1+x^2)^{\frac{5}{4}}\sin5x}{\sqrt{1-x^2}}~dx$

$=\int_0^1\dfrac{x^4(1+x^2)^{\frac{1}{4}}\sin5x}{\sqrt{1-x^2}}~d(x^2)-2\int_0^1\dfrac{(1+x^2)^{\frac{5}{4}}\sin5x}{\sqrt{1-x^2}}~d(x^2)$