# 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]** 

3 years, 5 months ago

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Problem 4

$\int_0^{\pi}\sqrt{1+4\sin^2 \frac x2-4\sin \frac x2} dx=?$

(Leave the answer in square roots and $\pi$-----No simplifications required..)

- 3 years, 5 months ago

Solution to problem 4:

$\sin(\dfrac{x}{2})\geq \dfrac{1}{2} \\ \dfrac{x}{2}\geq \arcsin(\dfrac{1}{2}) \\ \dfrac{x}{2}\geq \dfrac{\pi}{6} \\ x\geq \dfrac{\pi}{3} \\ Now, 1+4\sin^2\dfrac{x}{2}-4\sin\dfrac{x}{2} = \displaystyle (2\sin\dfrac{x}{2}-1)^2 \\ And \displaystyle \sqrt{1+4\sin^2\dfrac{x}{2}-4\sin\dfrac{x}{2}} =|2\sin\dfrac{x}{2}-1|$

So, the integral is : $\displaystyle \int_0^\pi |2\sin\dfrac{x}{2}-1| dx$ which can be split into two limits - $0 \ to \ \dfrac{\pi}{3} \ and \ \dfrac{\pi}{3} \ to \ \pi$ to remove the mod sign.

$\displaystyle \int_0^\dfrac{\pi}{3} (1-2\sin\dfrac{x}{2}) \ dx +\displaystyle \int_\dfrac{\pi}{3}^\pi (2\sin\dfrac{x}{2}-1) \ dx$

Now substituting the proper limits to this simple integral we get the answer as $4\sqrt3-4-\dfrac{\pi}{3}$

- 3 years, 5 months ago

PROBLEM 1:

$\int_a^{b} \frac{1}{x^2+ax+b} \mathrm{d}x$

Answer in terms of $a$ and $b$ only where $4b>a^2$.

This problem was solved by Nihar Mahajan.

- 3 years, 5 months ago

SOLUTION TO PROBLEM 1:

The integral can be written as $\displaystyle\int_{a}^{b} \dfrac{1}{\left(x+\dfrac{a}{2}\right)^2 + b-\dfrac{a^2}{4}} \ dx$ . Now substitute $y=x+\dfrac{a}{2}$ to get $\displaystyle\int_{3a/2}^{(2b+a)/2} \dfrac{1}{ y^2+ b-\dfrac{a^2}{4}} \ dy$. Now using $\int \dfrac{1}{x^2+a} = \dfrac{\tan^{-1}\left(\dfrac{x}{\sqrt{a}}\right)}{\sqrt{a}}+C$ , and substituting limits , the answer IS $\dfrac{2}{\sqrt{4b-a^2}} \tan^{-1}\left(\dfrac{2(b-a)(4b-a^2)}{2a^2+6ab+4b}\right)$ .

Partial credits to Vighnesh Shenoy to give me a start to this problem and Akshay for correcting my answer.

- 3 years, 5 months ago

PROBLEM 2:

$\int_0^1 \dfrac{\ln \dfrac{(x+a)^{x+a}}{(x+b)^{x+b}}}{(x+a)(x+b)\ln (x+a)\ln (x+b)}\ dx.$

This problem was solved by Samarth Agarwal.

- 3 years, 5 months ago

SOLUTION TO PROBLEM 2:

$\ln\dfrac{(x+a)^{x+a}}{(x+b)^{x+b}}=(x+a)\ln(x+a) - (x+b)\ln(x+b)$

$\dfrac{\ln\dfrac{(x+a)^{x+a}}{(x+b)^{x+b}}}{(x+a)(x+b)\ln(x+a)\ln(x+b)}=\dfrac{1}{(x+b)\ln(x+b)} - \dfrac{1}{(x+a)\ln(x+a)}$

now to evaluate $\displaystyle \int \dfrac{dx}{(x+k)\ln(x+k)}$

$\ln(x+k)=t$

$\dfrac{1}{x+k} dx =dt$

$\therefore the\ integral\ is\ \dfrac{dt}{t} = \ln(t) = \ln(x+t)$

Using this and putting limits we get the answer as :

$\ln\dfrac{\ln(1+b)\ln(a)}{\ln(1+a)\ln(b)}$

- 3 years, 5 months ago

PROBLEM 3:

$\displaystyle \int \dfrac{2x-1}{x^2-2x+10} \ dx$

This question was solved by Rishabh Cool and Akshay Yadav at almost same time.

- 3 years, 5 months ago

SOLUTION TO PROBLEM 3:

$\displaystyle \int \dfrac{2(x-1)+1}{x^2-2x+10} \ dx$

$=\displaystyle \int \dfrac{2(x-1)}{x^2-2x+10}+\dfrac{1}{(x-1)^2+9} \ dx$

$=\ln(x^2-2x+10)+\dfrac 13(\tan^{-1}(\frac{x-1}{3}))+C$

- 3 years, 5 months ago

Problem 5:

if $f(x)f(\dfrac{1}{x})=f(x)+f (\dfrac{1}{x})$ and$f(10)=1001$ , find $f(5)$.

P.S.: this is an easy problem to initiate calculus problems other than integration.

This problem was solved by Vighnesh Shenoy

- 3 years, 5 months ago

Let us consider f(x) is a constant c.
We get,
$c^{2} = 2c$
$c= 2$ or $c = 0$
Both of these do not satisfy,
$f(10) = 1001$
Now,
$f(x) = \dfrac{f\left(\dfrac{1}{x}\right)}{f\left(\dfrac{1}{x}\right) - 1}$
$\therefore \left( f(x)-1 \right) \cdot \left ( f\left(\dfrac{1}{x}\right) - 1 \right) = 1$
Let,
$f(x) = g(x) + 1 \rightarrow f\left(\dfrac{1}{x}\right) = g\left(\dfrac{1}{x}\right) + 1$
$\therefore g(x) \cdot g\left(\dfrac{1}{x}\right) = 1$
$g(x)$ is a polynomial of the type $\pm x^{n}$
$\therefore f(x) = 1 \pm x^{n}$
Substitute x = 10,
$1001 = 1 \pm 10^{n}$
$\pm 10^{n} = 1000$
$10^{n}$ can not be negative.
$\therefore 10^{n} = 1000 \rightarrow n = 3$
$f(x) = x^{3} + 1$
$f(5) = 125 + 1 = 126$

I remember our sir discussing this specific type of function once in class. I also feel this is more of algebra rather than calculus.

- 3 years, 5 months ago

PROBLEM 10:

If,

$S=-\cos^3(x) -\frac{1}{2}\cos^5(x)-\frac{1}{3}\cos^7(x)-\frac{1}{4}\cos^9(x)-...\infty$

Then find $\displaystyle \int_{\frac{\pi}{2}}^{\frac{5\pi}{2}} S \mathrm{d}x$.

This problem was solved by Samarth Agarwal but Vishnu Bhagyanath will post the next question.

- 3 years, 5 months ago

Area under the graph of $\cos^{2n-1} (x)$ from $k \text{ to } 2 \pi + k$ is zero as the negative half of the graph cancels out with the positive half.

Akshay's Approach:

Let me familiarize you with the series I used here,

$\ln(1-x^2)=-\sum_{n=1}^{\infty} \frac{x^{2n}}{n}$

You must take $\cos x$ common and then the integral would become,

$\displaystyle \int \cos(x)\ln(\sin^2 (x))\mathrm{d}x$

- 3 years, 5 months ago

- 3 years, 5 months ago

Problem 13

$I_n=\int_{-\pi}^{\pi}\left(\dfrac{\sin nx }{(1+\pi^x)\sin x}\right)\mathrm{d}x$ $n=0,1,2,....$

$I_{n+2}-I_n=k\times 100!, k\in \mathbf{Z}$

Find $k!$

This problem was solved by Nihar Mahajan and later by Vighnesh Shenoy

- 3 years, 5 months ago

SOLUTION TO PROBLEM 13:

Note that $I_{n+2} - I_n = \displaystyle\int_{-\pi}^{\pi} \dfrac{\sin (nx+2x) - \sin nx}{(1+\pi^{x})\sin x} \ dx = \displaystyle\int_{-\pi}^{\pi} \dfrac{2\sin x \cos ((n+1)x)}{(1+\pi^{x})\sin x} \ dx = \int_{-\pi}^{\pi} \dfrac{2 \cos ((n+1)x)}{(1+\pi^{x})} \ dx$

Now since the integrand is an even function we use the integration trick $\displaystyle\int_{-a}^{a} \dfrac{E(x)}{1+\pi^x} \ dx = \displaystyle\int_{0}^{a} E(x) \ dx$ :

$\int_{0}^{\pi} 2 \cos ((n+1)x) = \dfrac{2\sin(n+1)}{n+1} \bigg|_{0}^{\pi} = 0$

Hence , $k=0\Rightarrow k!=1$ .

- 3 years, 5 months ago

$I_{n} = \displaystyle \int_{-\pi}^{\pi} \dfrac{\sin(nx)dx}{(1+\pi^{x})\sin(x)} = \int_{0}^{\pi} \dfrac{\sin(nx)dx}{\sin(x)}$

$I_{n+2} - I_{n} = \displaystyle \int_{0}^{\pi} \dfrac{\sin((n+2)x) - \sin(nx) dx }{\sin x }$
$I_{n+2} - I_{n}= \displaystyle \int_{0}^{\pi} \dfrac{2\sin(x) \cos((n+1)x)dx}{\sin x } = 2\int_{0}^{\pi} \cos((n+1)x) dx = 0$
$k = 0, k! = 1$

- 3 years, 5 months ago

PROBLEM 7:

Find all real numbers $x$ such that $\displaystyle\int_0^x t^2\sin (x-t)\ dt=x^2$

This problem has been solved by Akshay Yadav.

- 3 years, 5 months ago

SOLUTION TO PROBLEM 7:

$\int_0^{x} t^2 \sin(x-t) \mathrm{d}t=x^2$

$\int_0^{x} t^2[ \sin(x)\cos(t)-\sin(t)\cos(x)] \mathrm{d}t=x^2$

$\int_0^{x} t^2 \sin(x)\cos(t)\mathrm{d}t- \int_0^{x} t^2 \sin(t)\cos(x) \mathrm{d}t=x^2$

Applying linearity and using Integration by parts,

$\sin(x)\int_0^{x} t^2 \cos(t)\mathrm{d}t-\cos(x) \int_0^{x} t^2 \sin(t)\mathrm{d}t=x^2$

$\sin(x)[x^2 \sin (x) -2\sin(x)+2x\cos (x)]-\cos(x)[-x^2\cos(x)+2x\sin(x)+2\cos(x) -2]=x^2$

$x^2+2\cos(x)-2=x^2$

$\cos(x)=1$

$x=2n\pi \text{ } \forall n \in \mathrm{I}$

- 3 years, 5 months ago

PROBLEM 11:

$\displaystyle \int_{0}^{\infty} e^{-\sqrt{x}}\mathrm{d}x$

This problem was solved by Vighnesh Shenoy.

- 3 years, 5 months ago

$x = t^{2} \rightarrow dx = 2tdt$
$I = \displaystyle 2 \int_{0}^{\infty}te^{-t}dt$
$\displaystyle \int te^{-t}dt = -e^{-t}(t+1)$
$I = 2\left[ -e^{-t}(t+1) \right]_{0}^{\infty} = -2\left( \displaystyle \lim_{x \rightarrow \infty} e^{-t}(t+1) - 1 \right)$ = 2

I did not use the gamma function on purpose as it is prohibited.

- 3 years, 5 months ago

Find the general solution to the differential equation,
$\dfrac{dy}{dx} - 2y\tan(x )= \sin(2x)$

This problem was solved by Rishabh Cool.

- 3 years, 5 months ago

Integration factor =$e^{\int -2\tan x dx}=\cos^2 x$.

$y\cos^2 x=\int 2\sin x \cos^3 xdx$ Make substitution $\cos x=t$ to evaluate the integral such that $-\sin x dx=dt$ $\boxed{y\cos ^2 x=-\dfrac{\cos^4 x}{2}+C}$

- 3 years, 5 months ago

PROBLEM 14:

Suppose $a,b$ are real numbers such that $a+b=1$. Then prove that the minimum value of the integral $\displaystyle\int_{0}^{\pi}(a\sin x+b\sin 2x)^{2}\ dx$ is $\dfrac{\pi}{4}$ and it occurs at $a=b=\dfrac{1}{2}$. (Use Calculus only).

This problem was solved by Vighnesh Shenoy.

- 3 years, 5 months ago

$I = \displaystyle \int_{0}^{\pi} (a\sin x + b\sin 2x)^{2} dx$

$I = \displaystyle \int_{0}^{\dfrac{\pi}{2}} (a\sin x + b \sin 2x )^{2} + (a \sin x - b \sin 2x)^{2} dx$

$\therefore I = \displaystyle 2\int_{0}^{\dfrac{\pi}{2}} a^{2} \sin^{2}x + b^{2} \sin^{2}2x dx$
On the first integral use, $\displaystyle \int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) + f(2a-x) dx$
$I =\displaystyle 2\int_{0}^{\dfrac{\pi}{4}}a^{2}dx + 4\int_{0}^{\dfrac{\pi}{4}}b^{2} \sin^{2}2x dx$
Use that property again on the second integral,
$I = 2\dfrac{a^{2}\pi}{4} + 4\dfrac{b^{2}\pi}{8} = \pi \cdot \dfrac{a^{2} + b^{2}}{2} = \pi \times \dfrac{(a^{2} + (1-a)^{2})}{2}$
Differentiate with respect to a,
$\dfrac{dI}{da} = \pi \times \dfrac{2a -2(1-a)}{2} = 0 \rightarrow a = \dfrac{1}{2} = b$
Differentiate again with respect to a,
$\dfrac{d^{2}I}{da^{2}} =\pi \times 2 > 0$
Thus the value is minimum, and occurs at $a = b = \dfrac{1}{2}$
$I_{min} = \dfrac{\pi}{2} \times \left( \left(\dfrac{1}{2}\right)^{2} \right) \times 2 = \dfrac{\pi}{4}$

- 3 years, 5 months ago

I have few doubts, doubt killer:

1) How did you get second step from first?

2) How is $\displaystyle \int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) + f(2a-x) dx$ ?

BTW Nice solution , Post next problem vighu.

- 3 years, 5 months ago

In the first step I used the same property.
Proof :
$I = \displaystyle \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)dx + \int_{a}^{2a}f(x)dx$
In the second integral,
$2a - x = t, \rightarrow dx = - dt$
$\displaystyle \int_{a}^{2a}f(x)dx = -\int_{a}^{0}f(2a-t)dt = \int_{0}^{a}f(2a-t)dt = \int_{0}^{2a}f(2a-x)dx$
$\therefore I = \displaystyle \int_{0}^{a}f(x) + f(2a-x)dx$

- 3 years, 5 months ago

Thanks!

- 3 years, 5 months ago

PROBLEM 6:

Find the closed form of the indefinite integral,
$\displaystyle \int \dfrac{dx}{1-\cot x}$

This problem has been solved by Nihar Mahajan

- 3 years, 5 months ago

SOLUTION TO PROBLEM 6: When we convert $\cot x$ in terms of $\sin x$ and $\cos x$ , we get:

$\displaystyle \int \dfrac{dx}{1-\cot x} = \dfrac{1}{2} \int \dfrac{2\sin x}{\sin x - \cos x} \ dx=\dfrac{1}{2} \int \dfrac{\sin x + \cos x + \sin x - \cos x}{\sin x - \cos x} \ dx \\ = \dfrac{1}{2} \left[\int 1 \ dx + \int \dfrac{\sin x + \cos x}{\sin x - \cos x} \ dx \right]$

Now substituting $u=\sin x - \cos x \Rightarrow du = \cos x+\sin x \ dx$ , the above integral changes to:

$\dfrac{1}{2}\left[x-\int \dfrac{1}{u} \ du\right] = \dfrac{x-\ln(|\cos x - \sin x|)}{2} + C$

- 3 years, 5 months ago

PROBLEM 8:

Find the length of an arc of curve

$y=\frac{a}{2}\left(e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right)$

between $x=0$ and $x=a$.

Answer in terms of $a$ and $e$ in simplest form.

This problem has been solved by Nihar Mahajan.

- 3 years, 5 months ago

SOLUTION TO PROBLEM 8:

Note that $y=a\cosh\left(\dfrac{x}{a}\right) \Rightarrow \dfrac{dy}{dx} = \sinh\left(\dfrac{x}{a}\right)$ (Using chain rule) . Now,

$\sqrt{1+\left(\dfrac{dy}{dx}\right)^2} = \sqrt{1+\sinh^2\left(\dfrac{x}{a}\right)} = \left| \cosh\left(\dfrac{x}{a}\right) \right|$

Now length of the arc from $x=0$ to $x=a$ is given by the integral: $\displaystyle\int_{0}^{a} \cosh\left(\dfrac{x}{a}\right) \ dx$

Substitute $y=\dfrac{x}{a} \Rightarrow dx=a \ dy$ and changing limits , the integral becomes $a\left[\sinh(y)\right]\bigg|_{0}^{1} = a\sinh(1) = \dfrac{a(e^2-1)}{2e}$

- 3 years, 5 months ago

PROBLEM 9:

Find the minimum area of the region bounded by the curve $y=a^3x^2-a^4x$ and the line $y=x$ where $a>0$ .

This problem has been solved by Akshay Yadav.

- 3 years, 5 months ago

SOLUTION TO PROBLEM 9:

I won't be providing a rigorous solution to problem as it is very long however here is what I did,

I figured the point of intersection of the two curves in terms of $a$.

I transformed the two curves that the area we need to calculate remains positive, (I am unable to provide an image of graph because of LaTeX, perhaps some one can help me).

Then integration to find area and subtraction, you will get area as a function of $a$.

Differentiate it and find the global minima.

- 3 years, 5 months ago

PROBLEM 15:

Evaluate the following integral,
$\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{4a\sin^{2}x + 6b^{2}\cos^{2}x}{(a^{2}\sin^{2}x + b^{3}\cos^{2}x)^{2}}dx$

This problem has been solved by Adarsh Kumar.

- 3 years, 5 months ago

$\dfrac{I}{2}=\int_{0}^{\frac{\pi}{2}}\dfrac{2a\sin^2x}{(a^2\sin^2x+b^3\cos^2x)^2}dx+\int_{0}^{\frac{\pi}{2}}\dfrac{3b^2\cos^2x}{(a^2\sin^2x+b^3\cos^2x)^2}dx$$I_1(a)=\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{a^2\sin^2x+b^3\cos^2x}$$\Longrightarrow I_1'(a)=\int_{0}^{\frac{\pi}{2}}\dfrac{-dx}{(a^2\sin^2x+b^3\cos^2x)^2}\times 2a\sin^2x$,the same definition goes for $I_2(b)$,and hence $I_2'(b)=\int_0^{\frac{\pi}{2}}\dfrac{-dx}{(a^2\sin^2x+b^3\cos^2x)^2}\times 3b^2\cos^2x$,hence,$I_1'(a)+I_2'(b)=-\dfrac{I}{2}(*)$,evaluating $I_1(a)$ and $I_2(b)$ using $\tan$ and $\sec$ technique and using the fact that $\int\dfrac{dx}{x^2+a^2}=\dfrac{1}{a}\tan^{-1}{x}$,we finally get that,$I_1'(a)=-\dfrac{1}{a^2b^{\frac{3}{2}}}\dfrac{\pi}{2}\\ \text{and}\ I_{2}'(b)=-\dfrac{3}{2}\dfrac{1}{ab^{\frac{5}{2}}}\dfrac{\pi}{2}$.Substituting these values in $*$ we get that $I=\pi\dfrac{3a+2b}{2a^2b^{\frac{5}{2}}}$,it is done!

- 3 years, 5 months ago

Problem 17:

$f(x) = \sqrt{x - \sqrt{12ax-36a^{2}}}$

Find the value of $\displaystyle \int f(x)dx$

This problem was solved by Nihar Mahajan.

- 3 years, 5 months ago

SOLUTION TO PROBLEM 17:

Note that $f(x) = \sqrt{\left(\sqrt{x-3a}\right)^2 + \left(\sqrt{3a}\right)^2 - 2\sqrt{3ax-9a^2}} = \sqrt{(\sqrt{x-3a} - \sqrt{3a})^2} = \sqrt{x-3a}-\sqrt{3a}$.

Hence, $\displaystyle\int \sqrt{x-3a}-\sqrt{3a} = \dfrac{2(x-3a)^{3/2}}{3} - \sqrt{3a}x+C$ .

- 3 years, 5 months ago

PROBLEM 19:

Find the quadratic mean and the arithmetical mean of the function $y=A_1\sin (x)+A_3\sin (3x)$

Subsequently find the minimum value of quadratic mean as $A_1,A_3 \in \mathbb{R}$.

This question was solved by Hummus A.

- 3 years, 5 months ago

here's my solution

quadratic mean: first we square the function,we get

${ A }_{ 1 }^{ 2 }\sin ^{ 2 }{ x } +2{ A }_{ 1 }{ A }_{ 3 }\sin { (x) } \sin { (3x) } +{ A }_{ 3 }^{ 2 }\sin ^{ 2 }{ (3x) }$

since all the functions here are periodic,the RMS over all time is the RMS over one period of the function

so computing the RMS we get

$\sqrt { \frac { 1 }{ \pi } \displaystyle\int _{ 0 }^{ \pi }{ { A }_{ 1 }^{ 2 }\sin ^{ 2 }{ x } +2{ A }_{ 1 }{ A }_{ 3 }\sin { (x) } \sin { (3x) } +{ A }_{ 3 }^{ 2 }\sin ^{ 2 }{ (3x) } dx } } =\\ \\ \sqrt { \frac { 1 }{ \pi } ({ A }_{ 1 }^{ 2 }\displaystyle\int _{ 0 }^{ \pi }{ \sin ^{ 2 }{ x } dx } +2{ A }_{ 1 }{ A }_{ 3 }\displaystyle\int _{ 0 }^{ \pi }{ \sin { (x)\sin { (3x)dx } +{ A }_{ 3 }^{ 2 } } \displaystyle\int _{ 0 }^{ \pi }{ \sin ^{ 2 }{ (3x) } dx } }) } =\sqrt { \frac { { A }_{ 1 }^{ 2 } }{ 2 } +\frac { { A }_{ 3 }^{ 2 } }{ 2 } }$

the minimum value of the quadratic mean here is 0,when ${ A }_{ 1 }={ A }_{ 3 }=0$,otherwise it has no minimum

mean:

this is the integral of the function over the period divided by the the difference of the bounds,which is

$\Large\frac { \displaystyle\int _{ -\pi /2 }^{ 3\pi /2 }{ \sin { x } +\sin { 3x } dx } }{ 2\pi } =\frac { 0 }{ 2\pi } =0$

i'm not sure if this solution is error free since i was too lazy to get pen and paper to work on it and did it mostly mentally,so feel free to notify me about any errors :)

- 3 years, 4 months ago

PROBLEM 20

prove that

$\displaystyle\int _{ a }^{ b }{ \frac { x\quad dx }{ \sqrt { (x-a)(b-x) } } } =\frac { \pi (a+b) }{ 2 }$

This problem was solved first by Adarsh Kumar and then by others.

- 3 years, 4 months ago

Solution to problem 20:$\int_a^b\dfrac{x dx}{\sqrt{(x-a)(b-x)}}$$=\int_a^b\dfrac{(a+b-x)dx}{\sqrt{(x-a)(b-x)}}$$\Longrightarrow 2I=(a+b)\int_a^b(a+b)\dfrac{dx}{\sqrt{(x-a)(b-x)}}$.Now substituting $z=\dfrac{a+b}{2}-x$,and changing the limits accordingly,$2I=-(a+b)\int_{\dfrac{b-a}{2}}^{\dfrac{a-b}{2}}\dfrac{dz}{\sqrt{(\dfrac{b-a}{2}-z)(\dfrac{b-a}{2}+z)}}$$\int \dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\dfrac{x}{a}$,doing the calculations we get,$2I=-(a+b)(-\pi)\\ \Longrightarrow I=\dfrac{(a+b)\pi}{2}$.It is done!

- 3 years, 4 months ago

WOW!, 4 people at the same time

- 3 years, 4 months ago

Another way is to take sqrt(x-a)=t and solve the trivial integral

- 3 years, 4 months ago

You post the next question ;)

- 3 years, 4 months ago

First simplifying $\displaystyle \int_a^b \frac{1}{\sqrt{(x-a)(b-x)}} \mathrm{d}x$, I would be providing beginning steps only as its very long generalization.

$\displaystyle \int_a^b \frac{1}{\sqrt{xb-x^2-ab+ax}} \mathrm{d}x$

$\displaystyle \int_a^b \frac{1}{\sqrt{-\left(x-\frac{a+b}{2}\right)^2+\left(\frac{a-b}{2}\right)^2}} \mathrm{d}x$

$\arcsin \left(\frac{2x}{a-b}-\frac{a+b}{a-b}\right)|_a^b$

$\arcsin (1)-\arcsin (-1) \rightarrow \pi$

- 3 years, 4 months ago

SOLUTION TO PROBLEM 20:

Very beautiful problem! Firstly, using $\int_{a}^{b} f(x) \ dx = \int_{a}^{b} f(a+b-x) \ dx$ , we have:

$I=\int_{a}^{b} \dfrac{x}{\sqrt{(x-a)(b-x)}} \ dx = \int_{a}^{b} \dfrac{a+b-x}{\sqrt{(x-a)(b-x)}} \ dx$

Adding both we get $2I= \displaystyle\int_{a}^{b}\dfrac{a+b}{\sqrt{(x-a)(b-x)}} \ dx$ . Now we need to get rid of mutiple variable of denominator by introducing single new variable. So substitute $y=\dfrac{x-a}{b-a}$ and note that $1-y = \dfrac{b-x}{b-a}$ and $(b-a)dy=dx$ and the integral changes to: $I=\dfrac{a+b}{2} \displaystyle\int_{0}^{1} \dfrac{1}{\sqrt{y(1-y)}} \ dy$ which is simply the antiderivative for $\arcsin$. So we have the integral evaluated as $\dfrac{\pi(a+b)}{2}$ .

- 3 years, 4 months ago

First using property $\int_a^b f(x)dx=\int f(a+b-x)dx$ and then adding to simplify the integral as:

$I=\int_a^b \frac {(a+b- x)dx }{ \sqrt { (x-a)(b-x) } }$

$=\int_a^b\dfrac{(a+b)dx}{ \sqrt { (x-a)(b-x) } }-I$

$\implies I=\int_a^b \dfrac{(a+b)}{2}(\dfrac{dx}{ \sqrt { (x-a)(b-x) } })$

$=\int_a^b\dfrac{a+b}{2}(\dfrac{dx}{ \sqrt {\dfrac{(b-a)^2}{2}- (x-(\dfrac{(a+b)}{2}))^2} })$

Now using $\int \dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1} \dfrac xa$.

$I=\dfrac{(a+b)}{2}(\sin^{-1}(\dfrac{2x-(a+b)}{b-a}))|_a^b$

$\large I=\boxed{\dfrac{\pi(a+b)}{2}}$

- 3 years, 4 months ago

Problem 21 by Adarsh Kumar:$\int[(x+a)^{-3}(x+b)^{-5}]^{\frac{1}{4}}dx=?$

This problem was first solved by Harsh Shrivastva.

- 3 years, 4 months ago

Solution 21 by Harsh Shrivastava:

- 3 years, 4 months ago

Problem 22

$\displaystyle \dfrac{\int_{0}^{\pi} x\ln \sin x \ dx}{\int_{0}^{\pi} \ln\sin x \ dx} =\pi/2$

This problem was first solved by Rishabh Cool.

- 3 years, 4 months ago

$I=\dfrac{\overbrace{\int_{0}^{\pi} x\ln \sin x \ dx}^{\color{#D61F06}{I_1}}}{\underbrace{\int_{0}^{\pi} \ln\sin x \ dx}_{\color{#D61F06}{I_2}}}$

$I_1=\int_{0}^{\pi} x\ln \sin x \ dx$

$=\int_0^{\pi} (\pi-x)\ln \sin (\pi-x) \ dx$

$=\int_{0}^{\pi} \pi\ln \sin x \ dx-I_1$

$\implies I_1=\dfrac{\pi}{2}\int_{0}^{\pi} \ln \sin x \ dx=\dfrac{\pi}{2}I_2$

And thus :

$\dfrac{I_1}{I_2}=\dfrac{\pi}{2}=I$

Hence proved...

- 3 years, 4 months ago