1
$\begingroup$

Today I posted a question about the integral

$$\int\frac{(\cos{c x})^2}{a+bx}dx$$

for which a Taylor series can be built and the integral solved for the desired approximation.

Another term in the stiffness matrix has the integral:

$$\int\frac{x^2(\cos{c x})^2}{a+bx}dx$$

whose Taylor series cannot be used since it does not converge:

$$x^4 \left(\frac{b^2}{a^3}-\frac{c^2}{a}\right)-\frac{b x^3}{a^2}+x^6 \left(\frac{b^4}{a^5}-\frac{b^2 c^2}{a^3}+\frac{c^4}{3 a}\right)+x^5 \left(\frac{b c^2}{a^2}-\frac{b^3}{a^4}\right)+\frac{x^2}{a}+O[x]^7$$

In this case, which method could be used to calculate the integral?

$\endgroup$
  • $\begingroup$ Riemann sums, or whatever other corresponding object if you're using an alternative definition of definite integration. $\endgroup$ – David H Sep 25 '13 at 13:23
  • $\begingroup$ Your use of "stiffness matrix" suggests a numerical application, perhaps FEM. Typically a numerical quadrature scheme such as a trapezoid rule (or midpoint, or Gaussian quadrature) is used to get numerical evaluation of integrals. $\endgroup$ – hardmath Sep 25 '13 at 13:57
  • $\begingroup$ "has no exact integration" does not make much sense. $\endgroup$ – leonbloy Sep 25 '13 at 15:30
  • $\begingroup$ @leonbloy I meant "has no closed form solution"... $\endgroup$ – Saullo G. P. Castro Sep 25 '13 at 15:51
2
$\begingroup$

Maple finds a closed form for your antiderivative:

$$ 1/4\,{\frac {x\sin \left( 2\,cx \right) }{bc}}+1/8\,{\frac {\cos \left( 2\,cx \right) }{{c}^{2}b}}-1/4\,{\frac {a\sin \left( 2\,cx \right) }{c{b}^{2}}}+1/2\,{a}^{2}{\it Si} \left( 2\,cx+2\,{\frac {ac} {b}} \right) \sin \left( 2\,{\frac {ac}{b}} \right) {b}^{-3}\\+1/2\,{a}^ {2}{\it Ci} \left( 2\,cx+2\,{\frac {ac}{b}} \right) \cos \left( 2\,{ \frac {ac}{b}} \right) {b}^{-3}+1/4\,{\frac {{x}^{2}}{b}}-1/2\,{\frac {ax}{{b}^{2}}}+1/2\,{\frac {{a}^{2}\ln \left( bcx+ac \right) }{{b}^{3 }}} $$ where Si and Ci are the Sine-integral and Cosine-integral functions.

$\endgroup$
1
$\begingroup$

Assume $b\neq0$ for the key case.

Let $u=a+bx$ ,

Then $x=\dfrac{u-a}{b}$

$dx=\dfrac{du}{b}$

$\therefore\int\dfrac{\cos^2cx}{a+bx}dx$

$=\int\dfrac{1+\cos2cx}{2(a+bx)}dx$

$=\int\dfrac{1+\cos\dfrac{2c(u-a)}{b}}{2bu}du$

$=\int\dfrac{1}{2bu}du+\int\dfrac{1}{2bu}\cos\dfrac{2ac}{b}\cos\dfrac{2cu}{b}du+\int\dfrac{1}{2bu}\sin\dfrac{2ac}{b}\sin\dfrac{2cu}{b}du$

$=\int\dfrac{1}{2bu}du+\int\dfrac{1}{2bu}\cos\dfrac{2ac}{b}\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{(2n)!}\left(\dfrac{2cu}{b}\right)^{2n}~du+\int\dfrac{1}{2bu}\sin\dfrac{2ac}{b}\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{(2n+1)!}\left(\dfrac{2cu}{b}\right)^{2n+1}~du$

$=\int\dfrac{1}{2bu}du+\int\dfrac{1}{2bu}\cos\dfrac{2ac}{b}du+\int\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}c^{2n}u^{2n-1}}{b^{2n+1}(2n)!}\cos\dfrac{2ac}{b}du+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^nc^{2n+1}u^{2n}}{b^{2n+2}(2n+1)!}\sin\dfrac{2ac}{b}du$

$=\dfrac{\ln u}{2b}+\dfrac{\ln u}{2b}\cos\dfrac{2ac}{b}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}c^{2n}u^{2n}}{b^{2n+1}(2n)!(2n)}\cos\dfrac{2ac}{b}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^nc^{2n+1}u^{2n+1}}{b^{2n+2}(2n+1)!(2n+1)}\sin\dfrac{2ac}{b}+C$

$=\dfrac{\ln(a+bx)}{2b}+\dfrac{\ln(a+bx)}{2b}\cos\dfrac{2ac}{b}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^{n-1}c^{2n}(a+bx)^{2n}}{b^{2n+1}(2n)!n}\cos\dfrac{2ac}{b}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^nc^{2n+1}(a+bx)^{2n+1}}{b^{2n+2}(2n+1)!(2n+1)}\sin\dfrac{2ac}{b}+C$

$\therefore\int\dfrac{x^2\cos^2cx}{a+bx}dx$

$=\int\dfrac{x^2(1+\cos2cx)}{2(a+bx)}dx$

$=\int\dfrac{\left(\dfrac{u-a}{b}\right)^2\left(1+\cos\dfrac{2c(u-a)}{b}\right)}{2bu}du$

$=\int\dfrac{u^2-2au+a^2}{2b^3u}du+\int\dfrac{u^2-2au+a^2}{2b^3u}\cos\dfrac{2ac}{b}\cos\dfrac{2cu}{b}du+\int\dfrac{u^2-2au+a^2}{2b^3u}\sin\dfrac{2ac}{b}\sin\dfrac{2cu}{b}du$

$=\int\left(\dfrac{u}{2b^3}-\dfrac{a}{b^3}+\dfrac{a^2}{2b^3u}\right)du+\int\left(\dfrac{u}{2b^3}-\dfrac{a}{b^3}+\dfrac{a^2}{2b^3u}\right)\cos\dfrac{2ac}{b}\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{(2n)!}\left(\dfrac{2cu}{b}\right)^{2n}~du+\int\left(\dfrac{u}{2b^3}-\dfrac{a}{b^3}+\dfrac{a^2}{2b^3u}\right)\sin\dfrac{2ac}{b}\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{(2n+1)!}\left(\dfrac{2cu}{b}\right)^{2n+1}~du$

$=\int\left(\dfrac{u}{2b^3}-\dfrac{a}{b^3}+\dfrac{a^2}{2b^3u}\right)\left(1+\cos\dfrac{2ac}{b}\right)du+\int\left(\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}c^{2n}u^{2n+1}}{b^{2n+3}(2n)!}-\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^nac^{2n}u^{2n}}{b^{2n+3}(2n)!}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}a^2c^{2n}u^{2n-1}}{b^{2n+3}(2n)!}\right)\cos\dfrac{2ac}{b}du+\int\left(\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^nc^{2n+1}u^{2n+2}}{b^{2n+4}(2n+1)!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}ac^{2n+1}u^{2n+1}}{b^{2n+4}(2n+1)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^na^2c^{2n+1}u^{2n}}{b^{2n+4}(2n+1)!}\right)\sin\dfrac{2ac}{b}du$

$=\left(\dfrac{u^2}{4b^3}-\dfrac{au}{b^3}+\dfrac{a^2\ln u}{2b^3}\right)\left(1+\cos\dfrac{2ac}{b}\right)+\left(\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}c^{2n}u^{2n+2}}{b^{2n+3}(2n)!(2n+2)}-\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^nac^{2n}u^{2n+1}}{b^{2n+3}(2n+1)!}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}a^2c^{2n}u^{2n}}{b^{2n+3}(2n)!(2n)}\right)\cos\dfrac{2ac}{b}+\left(\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^nc^{2n+1}u^{2n+3}}{b^{2n+4}(2n+1)!(2n+3)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}ac^{2n+1}u^{2n+2}}{b^{2n+4}(2n+2)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^na^2c^{2n+1}u^{2n+1}}{b^{2n+4}(2n+1)!(2n+1)}\right)\sin\dfrac{2ac}{b}+C$

$=\left(\dfrac{(a+bx)^2}{4b^3}-\dfrac{a(a+bx)}{b^3}+\dfrac{a^2\ln(a+bx)}{2b^3}\right)\left(1+\cos\dfrac{2ac}{b}\right)+\left(\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^{n-1}c^{2n}(a+bx)^{2n+2}}{b^{2n+3}(2n)!(n+1)}-\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^nac^{2n}(a+bx)^{2n+1}}{b^{2n+3}(2n+1)!}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}a^2c^{2n}(a+bx)^{2n}}{b^{2n+3}(2n)!(2n)}\right)\cos\dfrac{2ac}{b}+\left(\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^nc^{2n+1}(a+bx)^{2n+3}}{b^{2n+4}(2n+1)!(2n+3)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}ac^{2n+1}(a+bx)^{2n+2}}{b^{2n+4}(2n+2)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^na^2c^{2n+1}(a+bx)^{2n+1}}{b^{2n+4}(2n+1)!(2n+1)}\right)\sin\dfrac{2ac}{b}+C$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.