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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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    It seems complicated. I wish $\sin 5x$ is $\cos 5x$.2011-12-26
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    Wolframalpha gives $0.295767$ for the first integral. And $1.36153$ for the second.2011-12-26
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    Is this a homework or research problem?2011-12-26
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    @Paul, it is $sin5x$. It is a homework. What method can be used for solving this integral?2011-12-26
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    Roin: What similar cases do you know how to solve? Which techniques do you know which you tried and failed? What indications did you receive from your teacher?2011-12-26
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    @DidierPiau, solving of this integral is required for my given problem.2011-12-26
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    Roin: Here is what I call a zero-content comment.2011-12-26
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    @Roin: The integrals appear to be difficult to evaluate in closed form. Perhaps "my given problem" can be solved in other ways. But possibly none of us can do that unless we know what that problem is.2011-12-26
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    Ok! Thank you for your assistance! However I should say that these integrals are regard the expansion of function$f(x)=(1+x^{2})^{\frac{1}{4}}sin5x$ in a series of legendre and chebyshev polynomials, so I think that they have one way to solve it.2011-12-27
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    Either you should accept the answer if you want or start a bounty (specifically saying what you did not find in these answers)2012-03-15
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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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Hint:

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

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

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

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

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

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

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

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

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