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:
1 2 3 4 5 6 7 

The comments will be easiest to follow if you sort by "Newest":
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Top NewestProblem 40 :
Evaluate \[\int_{0}^{1} \dfrac{5x^3+3x1}{(x^3+3x+1)^3} dx\] – Samuel Jones · 8 months ago
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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 \(1/25\) – Saarthak Marathe · 8 months ago
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\[ \begin{aligned}\int\frac{5x^3 + 3x  1}{\left(x^3 + 3x + 1\right)^3}\, dx &=\int\frac{5x^3 + 3x1}{\left(\sqrt{x}\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)\right)^3}\,dx\\&=\int\frac{5x^3 + 3x1}{x^{3/2}\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)^3}\, dx\\&=\int\frac{5x^{3/2}+3x^{1/2}x^{3/2}}{\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)^3}\, dx\\&=\int\frac{2\,d(x^{5/2}+3\sqrt{x}+x^{1/2})}{\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)^3}\\& \frac{1}{\left(x^{5/2}+3\sqrt{x}+x^{1/2}\right)^2}+C\\& \end{aligned} \]
\[ \implies \int_{0}^{1} \frac{5x^3 + 3x  1}{\left(x^3 + 3x + 1\right)^3} dx = \boxed{\dfrac{1}{25}} \] – Samuel Jones · 8 months ago
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– Saarthak Marathe · 8 months ago
Nice method.Log in to reply
Problem 41:
Evaluate
\[\displaystyle \int _{0}^{\pi} \frac{1}{ (5+\cos (x))^{\frac{1}{3}}} dx\] – Saarthak Marathe · 8 months ago
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@Saarthak Marathe Can you post your solution to this question? Thanks. – Samuel Jones · 7 months, 3 weeks ago
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\(I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { x } \sqrt { 1x } { (2+x) }^{ 1/3 } } } \)
I have posted this question on Brilliant hoping to get a less complicated solution to this question. – Saarthak Marathe · 7 months, 3 weeks ago
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@Saarthak Marathe That goes against the rules. Please post questions that you know the solution to from next time. – Samuel Jones · 7 months, 2 weeks ago
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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. – Samuel Jones · 7 months, 4 weeks ago
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– Aditya Sharma · 7 months, 4 weeks ago
Hmm was it changed? I mean he gave a problem which was there in brilliant and he posted another one which is this one.Log in to reply
@Aditya Sharma @Saarthak Marathe Earlier problem was to evaluate \(\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\sin^2 x +k^2 \cos^2 x) \ dx\) – Samuel Jones · 7 months, 4 weeks ago
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– Aditya Sharma · 7 months, 4 weeks ago
Ya. This was pointed out to be a problem on brilliant with a name "the one with log and trigonometry" (maybe). So he changed the problemLog in to reply
JEE contest? – Samuel Jones · 7 months, 4 weeks ago
Oh ok. Btw, do you know why alternate solutions have been deleted inLog in to reply
– Aditya Sharma · 7 months, 4 weeks ago
The contest which sandeep sir started? I thnik only particular solutions are kept for each problem.Log in to reply
– Samuel Jones · 7 months, 4 weeks ago
Why is that so? I think if a solution is entirely different from the accepted one, it shouldn't be deleted.Log in to reply
– Aditya Sharma · 7 months, 4 weeks ago
I do agree with u personally in that case, maybe the moderator wasn't too much careful while removing the solution. If sandeep sir or Calvin sir is informed they will rectify itLog in to reply
– Samuel Jones · 7 months, 4 weeks ago
I have informed them. Hope they'll resolve the issue.Log in to reply
@Saarthak Marathe should post the solution as the allotted time is over. – Aditya Sharma · 7 months, 4 weeks ago
Yeah andLog in to reply
– Aditya Sharma · 7 months, 4 weeks ago
I think the solution should be postedLog in to reply
@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 ! – Aditya Sharma · 7 months, 4 weeks ago
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– Akshay Yadav · 7 months, 4 weeks ago
Since Sarthak Marathe has exceeded the time limit, from this moment onwards anyone can post the new question.Log in to reply
– Aditya Kumar · 8 months ago
Can you verify the problem?Log in to reply
@Aditya Kumar The problem is correct. – Saarthak Marathe · 8 months ago
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Hey, Can you lift the bar on the level. I would also like to solve a few good problems – Vignesh S · 8 months ago
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Problem 46:
\(\displaystyle \int_{\pi}^{\pi} \dfrac{2x(1+sinx)}{1+{cos}^{2}x}dx \) – Saarthak Marathe · 7 months, 2 weeks ago
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Problem 43:
Evaluate
\[ \huge \int_{0}^{\frac{\pi}{2}} \dfrac{\ln (\cos x)}{\sin x} dx\] – Samuel Jones · 7 months, 3 weeks ago
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\(\displaystyle \text{I} = \int_{0}^{1} \dfrac{\log t}{1t^2} \ dt \)
\(\displaystyle = \sum_{n=0}^{\infty} \int_{0}^{1} t^{2n} \log t \ dt \quad \left(\text{Using infinite G.P.} \right)\)
\(\displaystyle \int_{0}^{1} t^{2n} \log t \ dt =  \dfrac{1}{(2n+1)^2} \) which can be shown using integration by parts many times.
\(\displaystyle \implies \text{I} =  \sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2} \)
\(\displaystyle =  \left(\sum_{n=1}^{\infty} \dfrac{1}{n^2}  \sum_{n=1}^{\infty} \dfrac{1}{(2n)^2} \right) \)
\(\displaystyle =  \dfrac{3}{4} \sum_{n=1}^{\infty} \dfrac{1}{n^2}\)
\(\displaystyle = \boxed{\dfrac{\pi^2}{8}}\) – Samuel Jones · 7 months, 3 weeks ago
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– Samarth Agarwal · 7 months, 3 weeks ago
Nice observation samuels (+1)!!Log in to reply
– Aditya Sharma · 7 months, 3 weeks ago
You should post the solution now.Log in to reply
@Aditya Sharma I've posted the solution, but you post the next problem, since you had solved it first. – Samuel Jones · 7 months, 3 weeks ago
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\(\text{Put cosx = t } \)
\(sinx dx=dt\implies dx = \frac{dt}{\sqrt{1t^2}}\)
\(\text{We have } J = \int_{0}^{1}\frac{ln(t)}{1t^2}\)
\(\text{Put ln(t) = u }\implies \frac{1}{t}dt=du\implies dt=tdu=e^u du\)
\(J = \int_{\infty}^{0} \frac{xe^x}{1e^{2x}}dx = \frac{1}{2}\int_{\infty}^{0} \frac{x[(e^x+1)+(e^x1)]}{1e^{2x}}dx\)
\(2J = \int_{\infty}^{0} [\frac{x}{1e^x}  \frac{x}{e^x+1}]dx\)
\(2J = \color{red}{\underbrace{\int_{0}^{\infty} [\frac{x}{e^x1}}} + \color{blue}{\underbrace{\frac{x}{e^x+1}]dx}}\)
\(\text{In this step we are going to implement some of the special functions , their definitions are as below : }\)
\(\text{Polygarithm Function :}\)
\(\large Li_{s}(z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s1}}{\frac{e^x}{z}1}dx\)
\(\large Li_{s}(z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s1}}{\frac{e^x}{z}+1}dx\)
\(\text{The special values where s=2 & z=1,1 are : } Li_{2}(1) = \frac{(\pi)^2}{6} , Li_{2}(1) = \frac{(\pi)^2}{12}\)
\(2J = \color{red}{\underbrace{Li_{2}(1)}} + \color{blue}{\underbrace{Li_{2}(1)}}\implies 2J = \frac{(\pi)^2}{6}+\frac{(\pi)^2}{12}\implies \boxed{J=\frac{\pi^2}{8}}\)
\(\text{There might be an method to do this by elementary functions . Please post if anyone gets this solved. }\) – Aditya Sharma · 7 months, 3 weeks ago
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@Samuel Jones – Adarsh Kumar · 7 months, 3 weeks ago
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?Log in to reply
@Aditya Sharma @Adarsh Kumar Do you want me to write the solution, or do I wait until you guys figure out your mistake ? – Samuel Jones · 7 months, 3 weeks ago
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@Samuel Jones – Aditya Sharma · 7 months, 3 weeks ago
Let the alotted time be over, then post your solution. Lets see if we can make it in that timeLog in to reply
– Samuel Jones · 7 months, 3 weeks ago
Sure.Log in to reply
– Samuel Jones · 7 months, 3 weeks ago
I have solved without special functions.Log in to reply
– Aditya Sharma · 7 months, 3 weeks ago
Is it the answer ?Log in to reply
– Samuel Jones · 7 months, 3 weeks ago
No, the answer is \( \dfrac{\pi^2}{8}\)Log in to reply
– Aditya Sharma · 7 months, 3 weeks ago
Yeah I tried it , maybe i m wrong but are special functions required? I too have the same question as adarshLog in to reply
– Eddy Li · 7 months, 3 weeks ago
I saw your solution, but in the step \(\frac {ln(t)}{1t^2}\), you have forgotten a negative sign when substituting. Therefore your answer is positive rather than negative. I will give a solution without special functions though so just wait for a while.Log in to reply
– Aditya Sharma · 7 months, 3 weeks ago
Thanks! I will fix it, I wasn't sure where I have missed the sign.Log in to reply
\(\mathbf{Problem }\) \(\mathbf{ 42}\)
\(\text{Integrate : } \large \int_{0}^{1} \frac{x^4(1x)^4}{1+x^2}dx\) – Aditya Sharma · 7 months, 4 weeks ago
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\( \displaystyle \int_{0}^{1} \dfrac{x^4(1x)^4}{1+x^2} dx = \int_{0}^{1} \left( x^6 4x^5 +5x^4 4x^2  \dfrac{4}{x^2+1} +4 \right) dx\)
\( = \boxed{\dfrac{22}{7}  \pi} \)
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 \(\pi\) . – Samuel Jones · 7 months, 3 weeks ago
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– Adarsh Kumar · 7 months, 3 weeks ago
It is a JEE problem yeah?(This came in the year my sister wrote IIT :P)Log in to reply
– Aditya Sharma · 7 months, 3 weeks ago
Oh you identified :p no need to mention the year, yes it's a Jee Advanced problemLog in to reply
– Adarsh Kumar · 7 months, 3 weeks ago
haha I had seen it some time before!That is why I was able to identify,i won't mention don't worry!Log in to reply
– Aditya Sharma · 7 months, 3 weeks ago
Haha it is worth sharing IIT JEE problemsLog in to reply
– Harsh Shrivastava · 7 months, 3 weeks ago
Yeah.Log in to reply
– Adarsh Kumar · 7 months, 3 weeks ago
I just completed reading the solution from a book!:P So,i guess i would wait sometime,in case somebody else is posting.Log in to reply
– Harsh Shrivastava · 7 months, 3 weeks ago
Is the solution very hard?Log in to reply
– Adarsh Kumar · 7 months, 3 weeks ago
Nah,just requires a small trick!Log in to reply
– Aditya Sharma · 7 months, 3 weeks ago
Yeah basically the method is to decompose and expand smartlyLog in to reply
– Adarsh Kumar · 7 months, 3 weeks ago
Exactly,I will post the solution by night,if nobody else does,then we will solve whole night!If you are free !Log in to reply
532456:Adarsh Kumar . we will have a night maths party :) – Aditya Sharma · 7 months, 3 weeks ago
Ya ofcourse !! If it's about maths and brilliant nothing comes more imp than it @mentionLog in to reply
Problem 45:
\(\displaystyle \int { \frac { x }{ { (7x10{ x }^{ 2 }) }^{ 3/2 } } dx } \) – Saarthak Marathe · 7 months, 3 weeks ago
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\(I=\displaystyle \int { \frac { x }{ ({ (2x)(x5) })^{ 3/2 } } dx } \)
\(I= \displaystyle \int { \frac { x }{ \frac { { (x5) }^{ 3/2 } }{ { (2x) }^{ 3/2 } } { (2x) }^{ 3 } } dx } \)
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. – Saarthak Marathe · 7 months, 3 weeks ago
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@Akshay Yadav Should I post the solution to this problem ,as no one has posted one yet ? – Saarthak Marathe · 7 months, 3 weeks ago
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– Akshay Yadav · 7 months, 3 weeks ago
Go ahead, post the solution. The time limit has been already exceeded for anyone to solve it.Log in to reply
@Akshay Yadav Should I post the next question? – Saarthak Marathe · 7 months, 2 weeks ago
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\(\text{Problem 44 :}\)
\(\large \int \frac{dx}{\sec^2x+\tan^2x}\) – Aditya Sharma · 7 months, 3 weeks ago
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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,
\(\sqrt { 2 } { \tan }^{ 1 }(\sqrt { 2 } \tan x)x+C \) – Saarthak Marathe · 7 months, 3 weeks ago
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– Aditya Sharma · 7 months, 3 weeks ago
Absolutely , Post the next problemLog in to reply