MHB How Can I Simplify Real Integration Calculations?

[aruwin]

Hi.
Is there a short way to calculate real integration? I tried it but it looks so tedious. I attached the question so please refer to that. I am stuck with no.1, by the way.
Here is my attempt:

∫_0^(2π) dx/(13 - 5*sin(x))


let u = tan(x/2) which means x/2 = arctan(u) which means x = 2*arctan(u);
let du = sec^2(x/2)/2 dx

∫ 2*sec^2(x/2) dx/[2*sec^2(x/2)*(13 - 5*sin(x))]

∫ 2 du/[sec^2(2*arctan(u)/2)*(13 - 5*sin(2*arctan(u)))]

∫ 2*cos^2(arctan(u)) du/[13 - 5*sin(2*arctan(u))]

∫ 2*cos^2(arctan(u)) du/[13 - 5*2*sin(arctan(u))*cos(arctan(u))]

adjacent = 1; opposite = u; hypotenuse = sqrt((u^2) + 1)

∫ 2*[(1/sqrt((u^2) + 1))^2] du / [13 - 10*(u/sqrt((u^2) + 1))*1/sqrt((u^2) + 1)]

∫ [2/((u^2) + 1)] du / [13 - (10u/((u^2) + 1))]

∫ 2 du / {[13 - (10u/((u^2) + 1))]*[(u^2) + 1]}

∫ 2 du / {13*[(u^2) + (1)] - (10u)}

∫ 2 du / [13*(u^2) + (13) - (10u)]

∫ 2 du / [13*(u^2) - (10u) + (13)]

∫ (2/13) du / [(u^2) - (10/13*u) + (1)]

∫ (2/13) du / [(u^2) - (10/13*u) + (1)]

p = (u^2) - (10/13*u) + 1

p + (25/169) = (u^2) - 10/13*u + 1 + (25/169)

p + (25/169) = ((u - 5/13)^2) + 1

p = ((u - 5/13)^2) + (144/169)

∫ (2/13) du / {[(u - (5/13))^2] + [(12/13)^2]}

b = u - 5/13; db = du

∫ (2/13) db / [(b^2) + ((12/13)^2)]

a = 12/13; b = 12/13*tan(t); db = 12/13*sec^2(t) dt

∫ (2/13) [12/13*sec^2(t) dt] / [((12/13)^2)*tan^2(t) + ((12/13)^2)]

∫ ((13/12)^2) * (2/13) * (12/13) * sec^2(t) dt /[tan^2(t) + 1]

∫ ((13/12)^2) * (2/13) * (12/13) dt = ∫ dt/6 = t/6

b = 12/13*tan(t) as t = arctan(13b/12)

arctan(13b/12)/6 as b = u - 5/13

arctan[13*(u - 5/13)/12]/6 = 1/6*arctan((13u - 5)/12) as u = tan(x/2) as x = 2*arctan(u)

NOT SURE WHAT THE FINAL ANSWER IS.


2) ∫_-∞^∞ dx/((x^2) + 4)^2

a = 2; x = 2*tan(t); dx = 2*sec^2(t) dt;

∫ 2*sec^2(t) dt/(4*tan^2(t) + 4)^2

1/8 * ∫ sec^2(t) dt/(sec^4(t))

1/8 * ∫ cos^2(t) dt

1/8 * ∫ (1 - sin^2(t)) dt

u = sin(t); du = cos(t) dt; dv = -sin(t); v = cos(t)

t + sin(t)*cos(t) - ∫cos^2(t) = ∫cos^2(t)

t + sin(t)*cos(t) = 2*∫cos^2(t)

[t + sin(t)*cos(t)]/16

x/2 = tan(t) as t = arctan(x/2)

[arctan(x/2) + sin(arctan(x/2))*cos(arctan(x/2))]/16

adjacent = x; opposite = 2; hypotenuse = sqrt((x^2) + 4)

[arctan(x/2) + (2/sqrt((x^2) + 4))*(x/sqrt((x^2) + 4))]/16

[arctan(x/2) + (2x/((x^2) + 4))]/16

limit x->∞ [arctan(x/2) + (2x/((x^2) + 4))]/16

limit x->∞ [(π/2) + 2/(2x)]/16 = π/32

limit x->-∞ [arctan(x/2) + (2x/((x^2) + 4))]/16

limit x->-∞ [(π/2) + 2/(2x)]/16 = -π/32

Final answer is π/16

 

[Prove It]

2) ∫_-∞^∞ dx/((x^2) + 4)^2

a = 2; x = 2*tan(t); dx = 2*sec^2(t) dt;

∫ 2*sec^2(t) dt/(4*tan^2(t) + 4)^2

1/8 * ∫ sec^2(t) dt/(sec^4(t))

1/8 * ∫ cos^2(t) dt

I agree with you up to here, now $\displaystyle \begin{align*} \cos^2{(t)} \equiv \frac{1}{2} + \frac{1}{2}\cos{(2t)} \end{align*}$ so

$\displaystyle \begin{align*} \int{ \cos^2{(t)}\,\mathrm{d}t} &= \int{ \frac{1}{2} + \frac{1}{2}\cos{(2t)}\,\mathrm{d}t} \\ &= \frac{1}{2}t + \frac{1}{4}\sin{(2t)} + C \\ &= \frac{1}{2}t + \frac{1}{2}\sin{(t)}\cos{(t)} \\ &= \frac{1}{2}t + \frac{1}{2}\tan{(t)}\cos^2{(t)} + C \\ &= \frac{1}{2}t + \frac{\tan{(t)}}{2\sec^2{(t)}} + C \\ &= \frac{1}{2}t + \frac{\tan{(t)}}{2 \left[ 1 + \tan^2{(t)} \right] } + C \\ &= \frac{1}{2}\arctan{ \left( \frac{x}{2} \right) } + \frac{\frac{x}{2}}{2 \left[ 1 + \left( \frac{x}{2} \right) ^2 \right] } + C \\ &= \frac{1}{2}\arctan{ \left( \frac{x}{2} \right) } + \frac{\frac{x}{2}}{ 2 \left( 1 + \frac{x^2}{4} \right) } +C \\ &= \frac{1}{2}\arctan{ \left( \frac{x}{2} \right) } + \frac{x}{4 + x^2} + C \end{align*}$

Anyway, since the integrand is an even function, that means

$\displaystyle \begin{align*} \int_{-\infty}^{\infty}{ \frac{\mathrm{d}x}{ \left( x^2 + 4 \right) ^2 } } &= 2\int_0^{\infty}{ \frac{\mathrm{d}x}{ \left( x^2 + 4 \right) ^2 } } \\ &= 2 \lim_{\epsilon \to \infty}{ \left[ \frac{1}{2}\arctan{ \left( \frac{x}{2} \right) } + \frac{x}{4 + x^2} \right] _0^{\epsilon} } \\ &= \frac{\pi}{2} \end{align*}$

 

[aruwin]

I agree with you up to here, now $\displaystyle \begin{align*} \cos^2{(t)} \equiv \frac{1}{2} + \frac{1}{2}\cos{(2t)} \end{align*}$ so

$\displaystyle \begin{align*} \int{ \cos^2{(t)}\,\mathrm{d}t} &= \int{ \frac{1}{2} + \frac{1}{2}\cos{(2t)}\,\mathrm{d}t} \\ &= \frac{1}{2}t + \frac{1}{4}\sin{(2t)} + C \\ &= \frac{1}{2}t + \frac{1}{2}\sin{(t)}\cos{(t)} \\ &= \frac{1}{2}t + \frac{1}{2}\tan{(t)}\cos^2{(t)} + C \\ &= \frac{1}{2}t + \frac{\tan{(t)}}{2\sec^2{(t)}} + C \\ &= \frac{1}{2}t + \frac{\tan{(t)}}{2 \left[ 1 + \tan^2{(t)} \right] } + C \\ &= \frac{1}{2}\arctan{ \left( \frac{x}{2} \right) } + \frac{\frac{x}{2}}{2 \left[ 1 + \left( \frac{x}{2} \right) ^2 \right] } + C \\ &= \frac{1}{2}\arctan{ \left( \frac{x}{2} \right) } + \frac{\frac{x}{2}}{ 2 \left( 1 + \frac{x^2}{4} \right) } +C \\ &= \frac{1}{2}\arctan{ \left( \frac{x}{2} \right) } + \frac{x}{4 + x^2} + C \end{align*}$

Anyway, since the integrand is an even function, that means

$\displaystyle \begin{align*} \int_{-\infty}^{\infty}{ \frac{\mathrm{d}x}{ \left( x^2 + 4 \right) ^2 } } &= 2\int_0^{\infty}{ \frac{\mathrm{d}x}{ \left( x^2 + 4 \right) ^2 } } \\ &= 2 \lim_{\epsilon \to \infty}{ \left[ \frac{1}{2}\arctan{ \left( \frac{x}{2} \right) } + \frac{x}{4 + x^2} \right] _0^{\epsilon} } \\ &= \frac{\pi}{2} \end{align*}$

What about no. 1? What is the final answer?

 

[Prove It]

What about no. 1? What is the final answer?

You're welcome >_<

 

[aruwin]

You're welcome >_<

Opps, sorry sorry! Thank you! I was so desperate about the answer that I forgot my manners. Forgive me.

 

[Prove It]

Hi.
Is there a short way to calculate real integration? I tried it but it looks so tedious. I attached the question so please refer to that. I am stuck with no.1, by the way.
Here is my attempt:

∫_0^(2π) dx/(13 - 5*sin(x))


let u = tan(x/2) which means x/2 = arctan(u) which means x = 2*arctan(u);
let du = sec^2(x/2)/2 dx

∫ 2*sec^2(x/2) dx/[2*sec^2(x/2)*(13 - 5*sin(x))]

∫ 2 du/[sec^2(2*arctan(u)/2)*(13 - 5*sin(2*arctan(u)))]

∫ 2*cos^2(arctan(u)) du/[13 - 5*sin(2*arctan(u))]

∫ 2*cos^2(arctan(u)) du/[13 - 5*2*sin(arctan(u))*cos(arctan(u))]

adjacent = 1; opposite = u; hypotenuse = sqrt((u^2) + 1)

∫ 2*[(1/sqrt((u^2) + 1))^2] du / [13 - 10*(u/sqrt((u^2) + 1))*1/sqrt((u^2) + 1)]

∫ [2/((u^2) + 1)] du / [13 - (10u/((u^2) + 1))]

∫ 2 du / {[13 - (10u/((u^2) + 1))]*[(u^2) + 1]}

∫ 2 du / {13*[(u^2) + (1)] - (10u)}

∫ 2 du / [13*(u^2) + (13) - (10u)]

∫ 2 du / [13*(u^2) - (10u) + (13)]

∫ (2/13) du / [(u^2) - (10/13*u) + (1)]

∫ (2/13) du / [(u^2) - (10/13*u) + (1)]

p = (u^2) - (10/13*u) + 1

p + (25/169) = (u^2) - 10/13*u + 1 + (25/169)

p + (25/169) = ((u - 5/13)^2) + 1

p = ((u - 5/13)^2) + (144/169)

∫ (2/13) du / {[(u - (5/13))^2] + [(12/13)^2]}

b = u - 5/13; db = du

∫ (2/13) db / [(b^2) + ((12/13)^2)]

a = 12/13; b = 12/13*tan(t); db = 12/13*sec^2(t) dt

∫ (2/13) [12/13*sec^2(t) dt] / [((12/13)^2)*tan^2(t) + ((12/13)^2)]

∫ ((13/12)^2) * (2/13) * (12/13) * sec^2(t) dt /[tan^2(t) + 1]

∫ ((13/12)^2) * (2/13) * (12/13) dt = ∫ dt/6 = t/6

b = 12/13*tan(t) as t = arctan(13b/12)

arctan(13b/12)/6 as b = u - 5/13

arctan[13*(u - 5/13)/12]/6 = 1/6*arctan((13u - 5)/12) as u = tan(x/2) as x = 2*arctan(u)

NOT SURE WHAT THE FINAL ANSWER IS.


2) ∫_-∞^∞ dx/((x^2) + 4)^2

a = 2; x = 2*tan(t); dx = 2*sec^2(t) dt;

∫ 2*sec^2(t) dt/(4*tan^2(t) + 4)^2

1/8 * ∫ sec^2(t) dt/(sec^4(t))

1/8 * ∫ cos^2(t) dt

1/8 * ∫ (1 - sin^2(t)) dt

u = sin(t); du = cos(t) dt; dv = -sin(t); v = cos(t)

t + sin(t)*cos(t) - ∫cos^2(t) = ∫cos^2(t)

t + sin(t)*cos(t) = 2*∫cos^2(t)

[t + sin(t)*cos(t)]/16

x/2 = tan(t) as t = arctan(x/2)

[arctan(x/2) + sin(arctan(x/2))*cos(arctan(x/2))]/16

adjacent = x; opposite = 2; hypotenuse = sqrt((x^2) + 4)

[arctan(x/2) + (2/sqrt((x^2) + 4))*(x/sqrt((x^2) + 4))]/16

[arctan(x/2) + (2x/((x^2) + 4))]/16

limit x->∞ [arctan(x/2) + (2x/((x^2) + 4))]/16

limit x->∞ [(π/2) + 2/(2x)]/16 = π/32

limit x->-∞ [arctan(x/2) + (2x/((x^2) + 4))]/16

limit x->-∞ [(π/2) + 2/(2x)]/16 = -π/32

Final answer is π/16

If you're going to make the substitution $\displaystyle \begin{align*} u = \tan{ \left( \frac{x}{2} \right) } \implies \mathrm{d}u = \frac{1}{2}\sec^2{ \left( \frac{x}{2} \right) } \end{align*}$ then

$\displaystyle \begin{align*} \sin{(x)} &= 2\sin{ \left( \frac{x}{2} \right) } \cos{ \left( \frac{x}{2} \right) } \\ &= 2 \tan{ \left( \frac{x}{2} \right) } \cos^2{ \left( \frac{x}{2} \right) } \\ &= \frac{2\tan{ \left( \frac{x}{2} \right) } }{\sec^2{ \left( \frac{x}{2} \right) } } \\ &= \frac{2\tan{ \left( \frac{x}{2} \right) } }{ 1 + \tan^2{ \left( \frac{x}{2}\right) } } \\ &= \frac{2u}{1 + u^2} \end{align*}$

and so the integral is

$\displaystyle \begin{align*} \int{ \frac{\mathrm{d}x}{13 - 5\sin{(x)}}} &= \int{ \frac{ \frac{1}{2}\sec^2{ \left( \frac{x}{2} \right) }\,\mathrm{d}x}{ \frac{1}{2}\sec^2{ \left( \frac{x}{2} \right) } \left[ 13 - 5\sin{(x)} \right] } } \\ &= \int{ \frac{\mathrm{d}u}{\frac{1}{2} \left[ 1 + \tan^2{ \left( \frac{x}{2} \right) } \right] \left\{ 13 - 5 \left[ \frac{2u}{1 + u^2} \right] \right\} } } \\ &= 2\int{ \frac{\mathrm{d}u}{\left( 1 + u^2 \right) \left( 13 - \frac{10u}{1 + u^2} \right) } } \\ &= 2\int{ \frac{\mathrm{d}u}{ 13u^2 - 10u + 13} } \\ &= \frac{2}{13} \int{ \frac{\mathrm{d}u}{u^2 - \frac{10}{13}u + 1 } } \\ &= \frac{2}{13} \int{ \frac{\mathrm{d}u}{ u^2 - \frac{10}{13}u + \left( -\frac{5}{13} \right) ^2 - \left( -\frac{5}{13} \right) ^2 + 1 } } \\ &= \frac{2}{13} \int{ \frac{\mathrm{d}u}{ \left( u - \frac{5}{13} \right) ^2 - \frac{25}{169} + \frac{169}{169}} } \\ &= \frac{2}{13} \int{ \frac{\mathrm{d}u}{ \left( u - \frac{5}{13} \right) ^2 + \frac{144}{169} } } \\ &= \frac{2}{13} \int{ \frac{\mathrm{d}u}{ \left( u - \frac{5}{13} \right) ^2 + \left( \frac{12}{13} \right) ^2 }} \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u - \frac{5}{13} = \frac{12}{13}\tan{ \left( \theta \right) } \implies \mathrm{d}u = \frac{12}{13} \sec^2{ \left( \theta \right) }\,\mathrm{d}\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{2}{13} \int{ \frac{\mathrm{d}u}{ \left( u -\frac{5}{13} \right) ^2 + \left( \frac{12}{13} \right) ^2 } } &= \frac{2}{13} \int{ \frac{ \frac{12}{13}\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta }{ \left[ \frac{12}{13} \tan{\left( \theta \right) } \right] ^2 + \left( \frac{12}{13} \right) ^2 } } \\ &= \frac{2}{13} \int{ \frac{ \frac{12}{13} \sec^2{ \left( \theta \right) } \,\mathrm{d}\theta }{ \left( \frac{12}{13} \right) ^2 \sec^2{ \left( \theta \right) } } } \\ &= \frac{1}{6} \int{ \mathrm{d}\theta } \\ &= \frac{1}{6}\theta + C \\ &= \frac{1}{6} \arctan{ \left( \frac{13u - 5}{12} \right) } + C \\ &= \frac{1}{6} \arctan{ \left[ \frac{13 \tan{ \left( \frac{x}{2} \right) } - 5 }{12} \right] } + C \end{align*}$

Thus

$\displaystyle \begin{align*} \int_0^{2\pi}{ \frac{\mathrm{d}x}{ 13 - 5\sin{(x)} } } &= \frac{1}{6} \left\{ \arctan{ \left[ \frac{13\tan{\left( \frac{x}{2} \right) } - 5}{12} \right] } \right\}_0^{2\pi} \\ &= \frac{1}{6} \left\{ \arctan{ \left[ \frac{13\tan{ \left( \pi \right) } - 5}{12} \right] } - \arctan{ \left[ \frac{13\tan{ (0) } - 5}{12} \right] } \right\} \\ &= \frac{1}{6} \left[ \arctan{ \left( -\frac{5}{12} \right) } - \arctan{ \left( -\frac{5}{12} \right) } \right] \\ &= 0 \end{align*}$

 

[chisigma]

It may be that some objection is to move to the 'solution' $\displaystyle \int_{0}^{2\ \pi} \frac{d \theta}{13 - 5\ \sin \theta} =0$. It is easy to see that in $[0,2\ \pi]$ is $\displaystyle \frac{1}{13 - 5\ \sin \theta} > 0$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

The procedure for solving the definite integral...

$\displaystyle I=\int_{0}^{2\ \pi} \frac{d \theta}{13 - 5\ \sin \theta}\ (1)$

... through the substition $\displaystyle u=\tan \frac{\theta}{2}$ is correct. Taking into account that for $\theta$ going from 0 to $2\ \pi$ u goes from $- \infty$ to $+ \infty$ we arrive at the integral...

$\displaystyle I = 2\ \int_{- \infty}^{+ \infty} \frac{d u}{13 - 10\ u + 13\ u^{2}} = \frac{1}{6}\ \lim_{l \rightarrow \infty} |\tan^{-1} \frac{13\ u - 5}{12}|_{- l}^{+ l} = \frac{\pi}{6}\ (2)$

Kind regards

$\chi$ $\sigma$

 

[Prove It]

The procedure for solving the definite integral...

$\displaystyle I=\int_{0}^{2\ \pi} \frac{d \theta}{13 - 5\ \sin \theta}\ (1)$

... through the substition $\displaystyle u=\tan \frac{\theta}{2}$ is correct. Taking into account that for $\theta$ going from 0 to $2\ \pi$ u goes from $- \infty$ to $+ \infty$ we arrive at the integral...

$\displaystyle I = 2\ \int_{- \infty}^{+ \infty} \frac{d u}{13 - 10\ u + 13\ u^{2}} = \frac{1}{6}\ \lim_{l \rightarrow \infty} |\tan^{-1} \frac{13\ u - 5}{12}|_{- l}^{+ l} = \frac{\pi}{6}\ (2)$

Kind regards

$\chi$ $\sigma$

But surely $\displaystyle \begin{align*} \tan{ \left( \frac{0}{2} \right) } = 0 \end{align*}$, not $\displaystyle \begin{align*} \to \infty \end{align*}$

 

[chisigma]

The procedure for solving the definite integral...

$\displaystyle I=\int_{0}^{2\ \pi} \frac{d \theta}{13 - 5\ \sin \theta}\ (1)$

... through the substition $\displaystyle u=\tan \frac{\theta}{2}$ is correct. Taking into account that for $\theta$ going from 0 to $2\ \pi$ u goes from $- \infty$ to $+ \infty$ we arrive at the integral...

$\displaystyle I = 2\ \int_{- \infty}^{+ \infty} \frac{d u}{13 - 10\ u + 13\ u^{2}} = \frac{1}{6}\ \lim_{l \rightarrow \infty} |\tan^{-1} \frac{13\ u - 5}{12}|_{- l}^{+ l} = \frac{\pi}{6}\ (2)$

Kind regards

$\chi$ $\sigma$

The result can be obtained in faster way using the general formula reported in...

http://mathhelpboards.com/calculus-10/integration-help-11145.html#post51926

... that I suggest memorizing ...

$\displaystyle \int \frac{dx}{a + b\ x + c\ x^{2}} = \frac{2}{\sqrt{\Delta}}\ \tan^{-1} \frac{b + 2\ c\ x}{\sqrt{\Delta}},\ \Delta = 4\ a\ c - b^{2}\ (1) $

Kind regards

$\chi$ $\sigma$

 

[Opalg]

But surely $\displaystyle \begin{align*} \tan{ \left( \frac{0}{2} \right) } = 0 \end{align*}$, not $\displaystyle \begin{align*} \to \infty \end{align*}$

As $\theta$ goes from $0$ to $2\pi$, $u = \tan\frac\theta2$ goes from $0$ to $\infty$ (in the first half of the interval) and then from $-\infty$ to $0$ (in the second half of the interval). So as $\theta$ goes through the entire interval, $u$ covers the whole real line from $-\infty$ to $\infty$.

 

[aruwin]

What about Residue Theorem? I think we can use that for this integration.

 

[Prove It]

What about Residue Theorem? I think we can use that for this integration.

Which function and contour would you choose?

 

[Opalg]

What about Residue Theorem? I think we can use that for this integration.

Yes, you can use that method. You need to make the substitution $z = e^{ix} = \cos x + i \sin x$. Then $dz = ie^{ix}dx = iz\,dz$ so that $dx = \frac {dz}{iz}.$ Also, $z^{-1} = \cos x - i \sin x$, so that $\sin x = \frac1{2i}{z-z^{-1}}.$ The integral then goes round the unit circle $C$ and is equal to $$\begin{aligned} \oint_C \frac1{13-\frac5{2i}(z-z^{-1})}\,\frac{dz}{iz}\ &= \oint_C \frac{2\,dz}{26iz - 5z(z-z^{-1})} \\ &= \oint_C \frac{-2\,dz}{5z^2 - 26iz - 5} \\ &= \oint_C \frac{-2\,dz}{(5z-i)(z-5i)}.\end{aligned}$$ By the residue theorem, that last integral is equal to $2\pi i$ times the residue at the only pole inside the contour, which is at $z = i/5.$ The residue there is $\dfrac{-2i}{24}$ and that gives the value of the integral as $\dfrac{\pi}6$ (in agreement with chisigma's result at comment #8 above).

You can also do the other integral $$\int_{-\infty}^{\infty} \frac{dx}{(x^2+4)^2}$$ by contour integration. This time, integrate $\dfrac1{(z^2+4)^2}$ around a contour consisting of the interval $[-R,R]$ of the $x$-axis followed by a semicircle of radius $R$ in the upper half-plane, and then let $R\to\infty$.