This is the third note to the season of Brilliant Sub Junior Calculus Contest (Season1).
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Hi everyone! This is the first season of Brilliant Sub Junior Calculus Contest. This contest is for beginners or intermediate ones who want to sharpen their skills of problemsolving in overall calculus.
The aim of this contest is to improve the skill of in the computation in all sorts of problem (of basic level) in calculus like integrals (both definite and indefinite), differentiation, limits, ecetra by learning from each other and of course to have fun!
Eligibility: People should fulfill either of the 2 following
17 years or below
Level 4 or below in Calculus
Eligible people here may participate in this contest.
The rules are as follows:
I will start by posting the first problem. If there is a user solves it, then they must post a new one.
You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.
Only make substantial comment that will contribute to the discussion.
Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.
If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.
The scope of questions is only computation of basic level problems in calculus.
It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.
You are also NOT allowed to post a solution using a contour integration or residue method.
Answer shouldn't contain any Special Function.
Please post your solution and your proposed problem in a single new thread.
Format your post is as follows:
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Top NewestProblem 40 :
Evaluate \[\int_{0}^{1} \dfrac{5x^3+3x1}{(x^3+3x+1)^3} dx\]
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\(I=\displaystyle \int { \frac { ({ x }^{ 3 }+3x+1)+6{ x }^{ 3 }+6x }{ { ({ x }^{ 3 }+3x+1) }^{ 3 } } dx } \)
This can be further written as,
\(I=\displaystyle \int { \frac { { ({ x }^{ 3 }+3x+1) }^{ 2 }+{ 2x({ x }^{ 3 }+3x+1) }({ 3x }^{ 2 }+3) }{ { ({ x }^{ 3 }+3x+1) }^{ 4 } } dx }\)
This is like the derivative form of \( \displaystyle \frac { x }{ { ({ x }^{ 3 } }+3x+1)^{ 2 }} \)
The integral simplifies to,
\(I=\displaystyle \frac { x }{ { ({ x }^{ 3 } }+3x+1)^{ 2 } } +constant\)
Now putting the limits \(0\) to \(1\) we get the value as \(1/25\)
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Just for enlightening, I'm posting my method too.
\[ \begin{aligned}\int\frac{5x^3 + 3x  1}{\left(x^3 + 3x + 1\right)^3}\, dx &=\int\frac{5x^3 + 3x1}{\left(\sqrt{x}\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)\right)^3}\,dx\\&=\int\frac{5x^3 + 3x1}{x^{3/2}\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)^3}\, dx\\&=\int\frac{5x^{3/2}+3x^{1/2}x^{3/2}}{\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)^3}\, dx\\&=\int\frac{2\,d(x^{5/2}+3\sqrt{x}+x^{1/2})}{\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)^3}\\& \frac{1}{\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)^2}+C\\& \end{aligned} \]
\[ \implies \int_{0}^{1} \frac{5x^3 + 3x  1}{\left(x^3 + 3x + 1\right)^3} dx = \boxed{\dfrac{1}{25}} \]
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Nice method.
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Problem 41:
Evaluate
\[\displaystyle \int _{0}^{\pi} \frac{1}{ (5+\cos (x))^{\frac{1}{3}}} dx\]
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Can you verify the problem?
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@Aditya Kumar The problem is correct.
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I think the solution should be posted
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@Akshay Yadav @Nihar Mahajan @Samuel Jones what you guyz suggest about continuing the contest as @Saarthak Marathe hasn't posted the solution yet . I think someone should post a new problem and keep it going !
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@Saarthak Marathe Why did you change the problem? I solved the original problem and came here to post the solution, only to find that it has been changed.
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Hmm was it changed? I mean he gave a problem which was there in brilliant and he posted another one which is this one.
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@Aditya Sharma @Saarthak Marathe Earlier problem was to evaluate \(\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\sin^2 x +k^2 \cos^2 x) \ dx\)
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JEE contest?
Oh ok. Btw, do you know why alternate solutions have been deleted inLog in to reply
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@Saarthak Marathe should post the solution as the allotted time is over.
Yeah andLog in to reply
@Saarthak Marathe Can you post your solution to this question? Thanks.
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To be honest,I don't know how to solve this one. When I first saw it,I thought it would be easy. But found out later that it is quite tough. It uses some kind of special function to solve. I managed to bring it to an algebraic form,
\(I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { x } \sqrt { 1x } { (2+x) }^{ 1/3 } } } \)
I have posted this question on Brilliant hoping to get a less complicated solution to this question.
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@Saarthak Marathe That goes against the rules. Please post questions that you know the solution to from next time.
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Hey, Can you lift the bar on the level. I would also like to solve a few good problems
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\(\mathbf{Problem }\) \(\mathbf{ 42}\)
\(\text{Integrate : } \large \int_{0}^{1} \frac{x^4(1x)^4}{1+x^2}dx\)
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Using partial fractions,
\( \displaystyle \int_{0}^{1} \dfrac{x^4(1x)^4}{1+x^2} dx = \int_{0}^{1} \left( x^6 4x^5 +5x^4 4x^2  \dfrac{4}{x^2+1} +4 \right) dx\)
\( = \boxed{\dfrac{22}{7}  \pi} \)
Interesting Note : Since the integrand is positive, the integral must also be positive, giving us \(\pi < \dfrac{22}{7}\) which is the widely used approximation for \(\pi\) .
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It is a JEE problem yeah?(This came in the year my sister wrote IIT :P)
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Oh you identified :p no need to mention the year, yes it's a Jee Advanced problem
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532456:Adarsh Kumar . we will have a night maths party :)
Ya ofcourse !! If it's about maths and brilliant nothing comes more imp than it @mentionLog in to reply
Problem 43:
Evaluate
\[ \huge \int_{0}^{\frac{\pi}{2}} \dfrac{\ln (\cos x)}{\sin x} dx\]
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Substitute \(\cos x = t\), the integral becomes
\(\displaystyle \text{I} = \int_{0}^{1} \dfrac{\log t}{1t^2} \ dt \)
\(\displaystyle = \sum_{n=0}^{\infty} \int_{0}^{1} t^{2n} \log t \ dt \quad \left(\text{Using infinite G.P.} \right)\)
\(\displaystyle \int_{0}^{1} t^{2n} \log t \ dt =  \dfrac{1}{(2n+1)^2} \) which can be shown using integration by parts many times.
\(\displaystyle \implies \text{I} =  \sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2} \)
\(\displaystyle =  \left(\sum_{n=1}^{\infty} \dfrac{1}{n^2}  \sum_{n=1}^{\infty} \dfrac{1}{(2n)^2} \right) \)
\(\displaystyle =  \dfrac{3}{4} \sum_{n=1}^{\infty} \dfrac{1}{n^2}\)
\(\displaystyle = \boxed{\dfrac{\pi^2}{8}}\)
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Nice observation samuels (+1)!!
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Hey,some of my friends told me that it is either quite hard or not possible to solve this without using special functions,does your solution also use any such functions? @Samuel Jones
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Yeah I tried it , maybe i m wrong but are special functions required? I too have the same question as adarsh
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I have solved without special functions.
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@Aditya Sharma @Adarsh Kumar Do you want me to write the solution, or do I wait until you guys figure out your mistake ?
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@Samuel Jones
Let the alotted time be over, then post your solution. Lets see if we can make it in that timeLog in to reply
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\(\text{Let the integral be J = } \int_{0}^{\frac{\pi}{2}} \frac{log(cosx)}{sinx}dx\)
\(\text{Put cosx = t } \)
\(sinx dx=dt\implies dx = \frac{dt}{\sqrt{1t^2}}\)
\(\text{We have } J = \int_{0}^{1}\frac{ln(t)}{1t^2}\)
\(\text{Put ln(t) = u }\implies \frac{1}{t}dt=du\implies dt=tdu=e^u du\)
\(J = \int_{\infty}^{0} \frac{xe^x}{1e^{2x}}dx = \frac{1}{2}\int_{\infty}^{0} \frac{x[(e^x+1)+(e^x1)]}{1e^{2x}}dx\)
\(2J = \int_{\infty}^{0} [\frac{x}{1e^x}  \frac{x}{e^x+1}]dx\)
\(2J = \color{red}{\underbrace{\int_{0}^{\infty} [\frac{x}{e^x1}}} + \color{blue}{\underbrace{\frac{x}{e^x+1}]dx}}\)
\(\text{In this step we are going to implement some of the special functions , their definitions are as below : }\)
\(\text{Polygarithm Function :}\)
\(\large Li_{s}(z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s1}}{\frac{e^x}{z}1}dx\)
\(\large Li_{s}(z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s1}}{\frac{e^x}{z}+1}dx\)
\(\text{The special values where s=2 & z=1,1 are : } Li_{2}(1) = \frac{(\pi)^2}{6} , Li_{2}(1) = \frac{(\pi)^2}{12}\)
\(2J = \color{red}{\underbrace{Li_{2}(1)}} + \color{blue}{\underbrace{Li_{2}(1)}}\implies 2J = \frac{(\pi)^2}{6}+\frac{(\pi)^2}{12}\implies \boxed{J=\frac{\pi^2}{8}}\)
\(\text{There might be an method to do this by elementary functions . Please post if anyone gets this solved. }\)
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You should post the solution now.
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@Aditya Sharma I've posted the solution, but you post the next problem, since you had solved it first.
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Problem 46:
\(\displaystyle \int_{\pi}^{\pi} \dfrac{2x(1+sinx)}{1+{cos}^{2}x}dx \)
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\(\text{Problem 44 :}\)
\(\large \int \frac{dx}{\sec^2x+\tan^2x}\)
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Take \( tanx=t\)
Then the integral becomes,
\( I=\displaystyle \int { \frac { dt }{ (1+{ t }^{ 2 })(1+{ 2t }^{ 2 }) } } \)
Now separate the terms,
\(I=\displaystyle 2\int { \frac { dt }{ (1+2{ t }^{ 2 }) } } \int { \frac { dt }{ ({ t }^{ 2 }+1) } } \)
which on integration and resubstituting the value \(t= \tan x\) comes out as,
\(\sqrt { 2 } { \tan }^{ 1 }(\sqrt { 2 } \tan x)x+C \)
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Absolutely , Post the next problem
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Problem 45:
\(\displaystyle \int { \frac { x }{ { (7x10{ x }^{ 2 }) }^{ 3/2 } } dx } \)
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Solution:
\(I=\displaystyle \int { \frac { x }{ ({ (2x)(x5) })^{ 3/2 } } dx } \)
\(I= \displaystyle \int { \frac { x }{ \frac { { (x5) }^{ 3/2 } }{ { (2x) }^{ 3/2 } } { (2x) }^{ 3 } } dx } \)
Take \(t=\frac{x5}{2x} \) which gives \(x=\frac{2t+5}{1+t} \)
\( I=\displaystyle \frac { 1 }{ 9 } \int { \frac { \frac { (2t+5) }{ (1+t) } }{ { t }^{ 3/2 }\frac { 1 }{ (1+t) } } } dt \)
This simplifies to ,
\(I=\displaystyle \frac { 2 }{ 9 } \int { { t }^{ 1/2 }dt+\frac { 5 }{ 9 } \int { { t }^{ 3/2 }dt } } \)
This on integration gives,
\(I=\displaystyle \frac { 4 }{ 9 } { t }^{ 1/2 }\frac { 10 }{ 9 } { t }^{ 1/2 } \)
Resubstituting \(t=\frac{x5}{2x}\) we get our answer.
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@Akshay Yadav Should I post the solution to this problem ,as no one has posted one yet ?
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Go ahead, post the solution. The time limit has been already exceeded for anyone to solve it.
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@Akshay Yadav Should I post the next question?
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