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# Brilliant Sub Junior Calculus Contest (Season-1) Note-3

This is the third note to the season of Brilliant Sub Junior Calculus Contest (Season-1).

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 problem-solving 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.

Format your post is as follows:

 1 2 3 4 5 6 7 **SOLUTION OF PROBLEM xxx (number of problem) :** **[Post your solution here]** **PROBLEM xxx (number of problem) :** **[Post your problem here]** 

1 year ago

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Problem 40 :

Evaluate $\int_{0}^{1} \dfrac{5x^3+3x-1}{(x^3+3x+1)^3} dx$ · 1 year ago

$$I=\displaystyle \int { \frac { -({ x }^{ 3 }+3x+1)+6{ x }^{ 3 }+6x }{ { ({ x }^{ 3 }+3x+1) }^{ 3 } } dx }$$

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 $$-1/25$$ · 1 year ago

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 + 3x-1}{\left(\sqrt{x}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)\right)^3}\,dx\\&=\int\frac{5x^3 + 3x-1}{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}

$\implies \int_{0}^{1} \frac{5x^3 + 3x - 1}{\left(x^3 + 3x + 1\right)^3} dx = \boxed{-\dfrac{1}{25}}$ · 1 year ago

Nice method. · 1 year ago

Problem 41:

Evaluate

$\displaystyle \int _{0}^{\pi} \frac{1}{ (5+\cos (x))^{\frac{1}{3}}} dx$ · 1 year ago

@Saarthak Marathe Can you post your solution to this question? Thanks. · 1 year ago

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,

$$I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { x } \sqrt { 1-x } { (2+x) }^{ 1/3 } } }$$

I have posted this question on Brilliant hoping to get a less complicated solution to this question. · 1 year ago

@Saarthak Marathe That goes against the rules. Please post questions that you know the solution to from next time. · 1 year ago

@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. · 1 year 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. · 1 year ago

@Aditya Sharma @Saarthak Marathe Earlier problem was to evaluate $$\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\sin^2 x +k^2 \cos^2 x) \ dx$$ · 1 year 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 problem · 1 year ago

Oh ok. Btw, do you know why alternate solutions have been deleted in JEE contest? · 1 year ago

The contest which sandeep sir started? I thnik only particular solutions are kept for each problem. · 1 year ago

Why is that so? I think if a solution is entirely different from the accepted one, it shouldn't be deleted. · 1 year 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 it · 1 year ago

I have informed them. Hope they'll resolve the issue. · 1 year ago

Yeah and @Saarthak Marathe should post the solution as the allotted time is over. · 1 year ago

I think the solution should be posted · 1 year ago

@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 ! · 1 year ago

Since Sarthak Marathe has exceeded the time limit, from this moment onwards anyone can post the new question. · 1 year ago

Can you verify the problem? · 1 year ago

@Aditya Kumar The problem is correct. · 1 year ago

Hey, Can you lift the bar on the level. I would also like to solve a few good problems · 1 year ago

Problem 46:

$$\displaystyle \int_{-\pi}^{\pi} \dfrac{2x(1+sinx)}{1+{cos}^{2}x}dx$$ · 1 year ago

Problem 43:

Evaluate

$\huge \int_{0}^{\frac{\pi}{2}} \dfrac{\ln (\cos x)}{\sin x} dx$ · 1 year ago

Substitute $$\cos x = t$$, the integral becomes

$$\displaystyle \text{I} = \int_{0}^{1} \dfrac{\log t}{1-t^2} \ dt$$

$$\displaystyle = \sum_{n=0}^{\infty} \int_{0}^{1} t^{2n} \log t \ dt \quad \left(\text{Using infinite G.P.} \right)$$

$$\displaystyle \int_{0}^{1} t^{2n} \log t \ dt = - \dfrac{1}{(2n+1)^2}$$ which can be shown using integration by parts many times.

$$\displaystyle \implies \text{I} = - \sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2}$$

$$\displaystyle = - \left(\sum_{n=1}^{\infty} \dfrac{1}{n^2} - \sum_{n=1}^{\infty} \dfrac{1}{(2n)^2} \right)$$

$$\displaystyle = - \dfrac{3}{4} \sum_{n=1}^{\infty} \dfrac{1}{n^2}$$

$$\displaystyle = \boxed{-\dfrac{\pi^2}{8}}$$ · 1 year ago

Nice observation samuels (+1)!! · 1 year ago

You should post the solution now. · 1 year ago

@Aditya Sharma I've posted the solution, but you post the next problem, since you had solved it first. · 1 year ago

$$\text{Let the integral be J = } \int_{0}^{\frac{\pi}{2}} \frac{log(cosx)}{sinx}dx$$

$$\text{Put cosx = t }$$

$$-sinx dx=dt\implies dx = -\frac{dt}{\sqrt{1-t^2}}$$

$$\text{We have } J = -\int_{0}^{1}\frac{ln(t)}{1-t^2}$$

$$\text{Put ln(t) = u }\implies \frac{1}{t}dt=du\implies dt=tdu=e^u du$$

$$J = -\int_{\infty}^{0} \frac{xe^x}{1-e^{2x}}dx = -\frac{1}{2}\int_{\infty}^{0} \frac{x[(e^x+1)+(e^x-1)]}{1-e^{2x}}dx$$

$$2J = -\int_{\infty}^{0} [\frac{x}{1-e^x} - \frac{x}{e^x+1}]dx$$

$$2J = -\color{red}{\underbrace{\int_{0}^{\infty} [\frac{x}{e^x-1}}} + \color{blue}{\underbrace{\frac{x}{e^x+1}]dx}}$$

$$\text{In this step we are going to implement some of the special functions , their definitions are as below : }$$

$$\text{Polygarithm Function :}$$

$$\large Li_{s}(z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s-1}}{\frac{e^x}{z}-1}dx$$

$$\large -Li_{s}(-z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s-1}}{\frac{e^x}{z}+1}dx$$

$$\text{The special values where s=2 & z=1,-1 are : } Li_{2}(1) = \frac{(\pi)^2}{6} , Li_{2}(-1) = -\frac{(\pi)^2}{12}$$

$$-2J = \color{red}{\underbrace{Li_{2}(1)}} + \color{blue}{\underbrace{-Li_{2}(-1)}}\implies -2J = \frac{(\pi)^2}{6}+\frac{(\pi)^2}{12}\implies \boxed{J=-\frac{\pi^2}{8}}$$

$$\text{There might be an method to do this by elementary functions . Please post if anyone gets this solved. }$$ · 1 year 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? @Samuel Jones · 1 year ago

@Aditya Sharma @Adarsh Kumar Do you want me to write the solution, or do I wait until you guys figure out your mistake ? · 1 year ago

Let the alotted time be over, then post your solution. Lets see if we can make it in that time @Samuel Jones · 1 year ago

Sure. · 1 year ago

I have solved without special functions. · 1 year ago

Is it the answer ? · 1 year ago

No, the answer is $$-\dfrac{\pi^2}{8}$$ · 1 year ago

Yeah I tried it , maybe i m wrong but are special functions required? I too have the same question as adarsh · 1 year ago

I saw your solution, but in the step $$\frac {ln(t)}{1-t^2}$$, 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. · 1 year ago

Thanks! I will fix it, I wasn't sure where I have missed the sign. · 1 year ago

$$\mathbf{Problem }$$ $$\mathbf{ 42}$$

$$\text{Integrate : } \large \int_{0}^{1} \frac{x^4(1-x)^4}{1+x^2}dx$$ · 1 year ago

Using partial fractions,

$$\displaystyle \int_{0}^{1} \dfrac{x^4(1-x)^4}{1+x^2} dx = \int_{0}^{1} \left( x^6 -4x^5 +5x^4 -4x^2 - \dfrac{4}{x^2+1} +4 \right) dx$$

$$= \boxed{\dfrac{22}{7} - \pi}$$

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 $$\pi$$ . · 1 year ago

It is a JEE problem yeah?(This came in the year my sister wrote IIT :P) · 1 year ago

Oh you identified :p no need to mention the year, yes it's a Jee Advanced problem · 1 year ago

haha I had seen it some time before!That is why I was able to identify,i won't mention don't worry! · 1 year ago

Haha it is worth sharing IIT JEE problems · 1 year ago

Yeah. · 1 year ago

I just completed reading the solution from a book!:P So,i guess i would wait sometime,in case somebody else is posting. · 1 year ago

Is the solution very hard? · 1 year ago

Nah,just requires a small trick! · 1 year ago

Yeah basically the method is to decompose and expand smartly · 1 year ago

Exactly,I will post the solution by night,if nobody else does,then we will solve whole night!If you are free ! · 1 year ago

Ya ofcourse !! If it's about maths and brilliant nothing comes more imp than it @mention532456:Adarsh Kumar . we will have a night maths party :-) · 1 year ago

Problem 45:

$$\displaystyle \int { \frac { x }{ { (7x-10-{ x }^{ 2 }) }^{ 3/2 } } dx }$$ · 1 year ago

Solution:

$$I=\displaystyle \int { \frac { x }{ ({ (2-x)(x-5) })^{ 3/2 } } dx }$$

$$I= \displaystyle \int { \frac { x }{ \frac { { (x-5) }^{ 3/2 } }{ { (2-x) }^{ 3/2 } } { (2-x) }^{ 3 } } dx }$$

Take $$t=\frac{x-5}{2-x}$$ 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 $$t=\frac{x-5}{2-x}$$ we get our answer. · 1 year ago

@Akshay Yadav Should I post the solution to this problem ,as no one has posted one yet ? · 1 year ago

Go ahead, post the solution. The time limit has been already exceeded for anyone to solve it. · 1 year ago

@Akshay Yadav Should I post the next question? · 1 year ago

$$\text{Problem 44 :}$$

$$\large \int \frac{dx}{\sec^2x+\tan^2x}$$ · 1 year ago

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,

$$\sqrt { 2 } { \tan }^{ -1 }(\sqrt { 2 } \tan x)-x+C$$ · 1 year ago