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Let $y = \frac\pi2 - x$. Then, the integral becomes $\displaystyle I = -\int \dfrac{dy}{(16 + 9\cos y)^2}$.

Using half angle tangent substitution, we have $t = \tan(\frac y2) , \cos y = \dfrac{1-t^2}{1+t^2}$ and $dy = \dfrac{2 \cdot dt}{1 + t^2}$. The integral transforms to

@Pi Han Goh
–
@Pi Han Goh how to solve this integration, is there any alternative methods also?
, $\int_{π}^{2π} \frac{(x^{2}+2) \cos x}{x^{3}} dx$
Thanks in advance

The integrand can be expressed as $\dfrac{1 + 9\cos x - 9\cos x}{(16 + 9\sin x)^2} = \dfrac{1 + 9\cos x}{(16 + 9\sin x)^2} - \dfrac{9\sin x}{(16 + 9\sin x)^2}.$

Split this into two integrals, for the first integral, let $y = \frac \pi2 - x$ then $t = \tan(\frac y2)$, and you can solve it like how I've written in my other comment. And use $u = 16 + 9\sin x$ to solve the second integral.

I suspect the indefinite integral is in the form of $A \cdot \dfrac{\sin x}x + B \cdot \dfrac{\sin x}{x^2} + C \cdot \dfrac{\cos x}{x} + D \cdot \dfrac{\cos x}{x^2},$ where $A,B,C,D$ are constants. Differentiating this expression with respect to $x$ should gives us back the integrand $\dfrac{(x^2 + 2)\cos x}{x^3}$.

Using quotient rule and comparing coefficients, we will get $(A,B,C,D) = (1,0,0,-1)$, thus $\int_\pi^{2\pi} \dfrac{ (x^2+2)\cos x}{x^3} \, dx = \left [ \dfrac{\sin x}x - \dfrac{\cos x}{x^2} \right ]_{\pi}^{2\pi} = \boxed{-\dfrac{5}{4\pi^2} } \approx-0.126651 .$

@Lil Doug
–
Wait... how are you getting a positive $\displaystyle \int \frac{\sin(x)}{x^2}$? It should be negative because $\displaystyle -\frac{1}{x^2}$

importmathdeffunction(x):return((x**2+2)*math.cos(x))/x**3n=2500b=6.28a=3.14deltaX=(b-a)/ndefmean(a,b):meanValue=(a+b)/2returnmeanValuexValue=[a]heightValue=[]foriinrange(0,n):xValue.append(a+deltaX)deltaX=deltaX+(b-a)/nforiinxValue:heightValue.append(function(mean(i,i+(b-a)/n)))delheightValue[-1]the_sum=sum(heightValue)definite_integral=the_sum*((b-a)/n)print('Definite integral evaluated by midpoint rule: '+str(definite_integral))

No, I looked. And you're right; they're not cancelling. My integration by parts has failed. We should work out $\displaystyle \int \frac{\sin(x)}{x^2}$

@Krishna Karthik it is not your integration by parts .
It is the mistake of this website https://www.integral-calculator.com/ who told you wrong method

@Lil Doug
–
Integral calculator? I worked most of this out by hand; I tried integrating by parts. I checked with integral calculator because I got stuck midway; it mistakingly said it would cancel out (integral calculator can be stupid sometimes...) so I copied an integration by parts method from them. Maybe there's some weird trig substitution or something that I'm missing. I'll try the problem soon.

Btw the my midpoint rule calculator seems to be coming up with a reasonable answer. This is actually a tricky problem... maybe integration by parts isn't the trick. Ask Pi Han Goh.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewest@Krishna Karthik Here it is, I'm going to bed now.

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@Steven Chase @Karan Chatrath , in case you guys are interested.

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Yeah; it's really impressive. Good job mate :)

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Bruh that's fuggin awesome!!!!

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@Krishna Karthik Do you know above line?

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Sorry? Yeah; you're going to bed. Cya.

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@Krishna Karthik it is the style of Steven Sir.

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@Krishna Karthik i forgot to write it in page. But this seems me more beautiful.

Conclusion:Everything happens for good

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Here's an alternative approach.

Let $y = \frac\pi2 - x$. Then, the integral becomes $\displaystyle I = -\int \dfrac{dy}{(16 + 9\cos y)^2}$.

Using half angle tangent substitution, we have $t = \tan(\frac y2) , \cos y = \dfrac{1-t^2}{1+t^2}$ and $dy = \dfrac{2 \cdot dt}{1 + t^2}$. The integral transforms to

$\begin{array} { r c l} I &=& \displaystyle -2 \int\dfrac1{\left (16 + 9 \left(\frac{1-t^2}{1+t^2} \right)\right)^2} \cdot \dfrac1{1+t^2} \, dt \\\phantom 0 \\ &=& \displaystyle -2 \int \dfrac{1+t^2}{ (16(1+t^2) + 9(1-t^2) )^2} \, dt \\ \phantom 0 \\ &=& \displaystyle -2 \int \dfrac{t^2 + 1}{ (7t^2 + 25 )^2} \, dt \\ \phantom0 \\ &=& \displaystyle -\frac27 \int \dfrac{7t^2 + 7}{ (7t^2 + 25 )^2} \, dt \\ \phantom0 \\ &=& \displaystyle -\frac27 \int \dfrac{7t^2 + 25 - 18}{ (7t^2 + 25 )^2} \, dt \\ \phantom0 \\ &=& \displaystyle -\frac27 \int \dfrac1{ 7t^2 + 25 } \, dt + \dfrac{36}{7} \int \dfrac1{(7t^2 + 25)^2} \, dt \\ \phantom0 \\ \end{array}$

For both integrals, apply the trigonometric substitution, $\sqrt7 t = 5 \tan u \Rightarrow \sqrt7 = 5\sec^2 u \cdot \dfrac{du}{dt}$.

$\begin{array} {r c l} I &=& \displaystyle -\frac27 \int \dfrac1{25\sec^2 u} \cdot \dfrac5{\sqrt7} \sec^2 u \, du + \dfrac{36}{7} \int \dfrac1{(25\sec^2 u)^2} \cdot \dfrac5{\sqrt7} \sec^2 u \, du \\ \phantom 0 \\ &=& \displaystyle -\frac2{35\sqrt7} \int du + \dfrac{36}{875\sqrt7} \int \cos^2 u \, du \\ \phantom 0 \\ &=& \displaystyle -\frac2{35\sqrt7} u + \dfrac{36}{875\sqrt7} \cdot \frac14 (\sin(2u) + 2u) +C \\ \phantom 0 \\ &=& \displaystyle -\frac{32}{875\sqrt7} u + \dfrac{9}{875\sqrt7} \sin(2u) + C \\ \phantom 0 \\ \end{array}$

What's left is to back-substitute. $u = \tan^{-1} \left( \frac{\sqrt7}5 t\right)= \tan^{-1} \left( \frac{\sqrt7}5 \tan\left( \frac y2\right)\right)= \tan^{-1} \left( \frac{\sqrt7}5 \tan\left( \frac {\pi /2 - x}2\right)\right)$. Finishing it off with $\tan\left( \frac {\pi /2 - x}2\right) = \dfrac{1 - \tan(\frac x2)}{1 + \tan(\frac x2)}$, we get:

$I = \displaystyle -\frac{32}{875\sqrt7} \tan^{-1} \left( \frac{\sqrt7}5 \tan\left( \dfrac{1 - \tan(\frac x2)}{1 + \tan(\frac x2)} \right)\right) + \dfrac{9}{875\sqrt7} \sin \left [ 2\left( \tan^{-1} \left( \frac{\sqrt7}5 \tan\left( \dfrac{1 - \tan(\frac x2)}{1 + \tan(\frac x2)} \right)\right) \right) \right] + C$

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@Pi Han Goh very very nice. Thanks for sharing.

By the way, do you know I am Neeraj . Due to some reason I have changed my name

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Now I know.

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@Pi Han Goh your integration skills are very nice.

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@Pi Han Goh

Nice one. I like it.

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@Pi Han Goh Btw have you ever participated in an Integration bee or an Integration competition? Holy shit... you're really good!

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No, but I'm aware of these competitions.

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@Pi Han Goh how to solve this integration, is there any alternative methods also? , $\int_{π}^{2π} \frac{(x^{2}+2) \cos x}{x^{3}} dx$

Thanks in advance

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@Pi Han Goh ok, I will post a note after one hour.

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Here's another method:

The integrand can be expressed as $\dfrac{1 + 9\cos x - 9\cos x}{(16 + 9\sin x)^2} = \dfrac{1 + 9\cos x}{(16 + 9\sin x)^2} - \dfrac{9\sin x}{(16 + 9\sin x)^2}.$

Split this into two integrals, for the first integral, let $y = \frac \pi2 - x$ then $t = \tan(\frac y2)$, and you can solve it like how I've written in my other comment. And use $u = 16 + 9\sin x$ to solve the second integral.

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@Pi Han Goh Thank you so much

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@Pi Han Goh @Krishna Karthik can you please help me in this integration $\int_{π}^{2π} \frac{(x^{2}+2) \cos x}{x^{3}} dx$

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@Lil Doug This one's actually quite easy. I'll post a solution soon.

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@Krishna Karthik please post it now.

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$\displaystyle \int \left( \frac{\cos(x)}{x} + \frac{2 \cos(x)}{x^3} \right) dx$

So, we can split it into two integrals, which we can do using

integration by parts.$\displaystyle \int \frac{\cos(x)}{x} dx + 2 \int \frac{\cos(x)}{x^3} dx$

So, we do integration by parts with the first half:

$\displaystyle u = \frac{1}{x}$, $\displaystyle dv = \cos(x)$

$\displaystyle du = - \frac{1}{x^2}$, $\displaystyle v = \sin(x)$

So, it can be rewritten as:

$\displaystyle I_1 = \frac{\sin(x)}{x} - \int \frac{\sin(x)}{x^2}$

Doing integration by parts for the second half, we get:

$\displaystyle I_2 = - \frac{\cos(x)}{2x^2} - \int \frac{\sin(x)}{x^2}$

Now simplifying, $\displaystyle \int \frac{\sin(x)}{x^2}$ cancels out, leaving:

$\displaystyle \frac{x \sin(x) - \cos(x)}{x^2} + C$

So substituting, you get:

$\displaystyle \boxed{ \frac{-5}{4 \pi^2}}$

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@Krishna Karthik how does it get cancel, both are negative, both will add up.

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$\displaystyle \int \frac{\sin(x)}{x^2}$

Oh wait... shit. I fucked up. So, let's try to calculate that integralLog in to reply

@Krishna Karthik what answer you are getting?

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I suspect the indefinite integral is in the form of $A \cdot \dfrac{\sin x}x + B \cdot \dfrac{\sin x}{x^2} + C \cdot \dfrac{\cos x}{x} + D \cdot \dfrac{\cos x}{x^2},$ where $A,B,C,D$ are constants. Differentiating this expression with respect to $x$ should gives us back the integrand $\dfrac{(x^2 + 2)\cos x}{x^3}$.

Using quotient rule and comparing coefficients, we will get $(A,B,C,D) = (1,0,0,-1)$, thus $\int_\pi^{2\pi} \dfrac{ (x^2+2)\cos x}{x^3} \, dx = \left [ \dfrac{\sin x}x - \dfrac{\cos x}{x^2} \right ]_{\pi}^{2\pi} = \boxed{-\dfrac{5}{4\pi^2} } \approx-0.126651 .$

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@Krishna Karthik i am getting $I_{1} =\frac{\sin x}{x}+\int \frac{\sin x }{x^{2}} dx$

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You missed out the $\displaystyle -\frac{\cos(x)}{x^2}$ term.

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@Krishna Karthik where bro, I am. Telling $I_{1}$

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$\displaystyle \int \frac{\sin(x)}{x^2}$? It should be negative because $\displaystyle -\frac{1}{x^2}$

Wait... how are you getting a positiveLog in to reply

Btw, numerically integrating, I'm getting -0.12779

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@Krishna Karthik share the code also.

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@Krishna Karthik share the code means share the code.

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@Lil Doug The code.

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Btw, you can shorten your code by doing this instead:

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@Krishna Karthik They are not cancelling bro

I think you have copied the solution and just paste it here.

Not taken a look on the solution

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No, I looked. And you're right; they're not cancelling. My integration by parts has failed. We should work out $\displaystyle \int \frac{\sin(x)}{x^2}$

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@Krishna Karthik it is not your integration by parts .

It is the mistake of this website https://www.integral-calculator.com/ who told you wrong method

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@Krishna Karthik. you code is also not working

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Btw the my midpoint rule calculator seems to be coming up with a reasonable answer. This is actually a tricky problem... maybe integration by parts isn't the trick. Ask Pi Han Goh.

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@Lil Doug I just posted a new note; it's a photo of the working code.

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