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Why start with $A = \cdots$ (yes, I know the answer to this question). But you seem to pull it out of thin air without any justification.

Say you used quotient rule, otherwise, the newbies can't follow this reasoning.

At least motivate why you want to work on the differentiation of $A$ first, by first explaining that the original integrand needs to be broken down (by considering a lesser power in the denominator).

Where did your $\sin \alpha =\cdots$ come from ? (I know the answer to this question too) It seems mystifying to the uninitiated.

At least make it explicit why you want convert the integrand in the form of $\frac1{t^2 + a^2}$ in the final step.

Handwriting issue: Some of your brackets does not properly encapsulate the required expressions.

Handwriting issue: Some of your fractions' horizontal line should be longer.

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.

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

@Talulah Riley
–
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.

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))

@Pi Han Goh I don't like to code in single-line style. I prefer readability as a programmer, personally.

Usually, I prefer to indent functions so I can see what is inside the function clearer. And btw thx for the suggestion of using it as just cos and pi rather than specifying the library.

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

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 $\int_\pi^{2\pi} \dfrac{ (x^2+2)\cos x}{x^3} \, dx = \left [ \dfrac{\sin x}x - \dfrac{\cos x}{x^2} \right ]_{\pi}^{2\pi} = \boxed{-\dfrac{5}{4\pi^2} } \approx-0.126651 .$

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.

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

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## Comments

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TopNewest@Pi Han Goh here above you can see

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A few pointers:

Why start with $A = \cdots$ (yes, I know the answer to this question). But you seem to pull it out of thin air without any justification.

Say you used quotient rule, otherwise, the newbies can't follow this reasoning.

At least motivate why you want to work on the differentiation of $A$ first, by first explaining that the original integrand needs to be broken down (by considering a lesser power in the denominator).

Where did your $\sin \alpha =\cdots$ come from ? (I know the answer to this question too) It seems mystifying to the uninitiated.

At least make it explicit why you want convert the integrand in the form of $\frac1{t^2 + a^2}$ in the final step.

Handwriting issue: Some of your brackets does not properly encapsulate the required expressions.

Handwriting issue: Some of your fractions' horizontal line should be longer.

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

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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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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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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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@Lil Doug The code.

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

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@Pi Han Goh I don't like to code in single-line style. I prefer readability as a programmer, personally.

Usually, I prefer to indent functions so I can see what is inside the function clearer. And btw thx for the suggestion of using it as just cos and pi rather than specifying the library.

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@Krishna Karthik No matter whatever you like, First upvote his alternative method of this note integration above.

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@Lil Doug Oh yeah; sorry I forgot to. You can't upvote subcomments, though, sadly.

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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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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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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

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

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@Krishna Karthik what answer you are getting?

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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

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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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 I have upvoted your solution here also, see above.

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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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@Pi Han Goh

Nice one. I like it.

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

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