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.

Please post your solution and your proposed problem in a single new thread.

Format your post is as follows:

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**SOLUTION OF PROBLEM xxx (number of problem) :**

**[Post your solution here]**

**PROBLEM xxx (number of problem) :**

**[Post your problem here]**

The comments will be easiest to follow if you sort by "Newest":

Note by Akshay Yadav
3 years, 7 months ago

No vote yet
1 vote

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Problem 4

0π1+4sin2x24sinx2dx=?\int_0^{\pi}\sqrt{1+4\sin^2 \frac x2-4\sin \frac x2} dx=?

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

Rishabh Jain - 3 years, 7 months ago

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Solution to problem 4:

First we start with the following inequality:

sin(x2)12x2arcsin(12)x2π6xπ3Now,1+4sin2x24sinx2=(2sinx21)2And1+4sin2x24sinx2=2sinx21\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 : 0π2sinx21dx \displaystyle \int_0^\pi |2\sin\dfrac{x}{2}-1| dx which can be split into two limits - 0 to π3 and π3 to π0 \ to \ \dfrac{\pi}{3} \ and \ \dfrac{\pi}{3} \ to \ \pi to remove the mod sign.

0π3(12sinx2) dx+π3π(2sinx21) dx\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 434π34\sqrt3-4-\dfrac{\pi}{3}

Samarth Agarwal - 3 years, 7 months ago

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PROBLEM 1:

ab1x2+ax+bdx\int_a^{b} \frac{1}{x^2+ax+b} \mathrm{d}x

Answer in terms of aa and bb only where 4b>a24b>a^2.

This problem was solved by Nihar Mahajan.

Akshay Yadav - 3 years, 7 months ago

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SOLUTION TO PROBLEM 1:

The integral can be written as ab1(x+a2)2+ba24 dx\displaystyle\int_{a}^{b} \dfrac{1}{\left(x+\dfrac{a}{2}\right)^2 + b-\dfrac{a^2}{4}} \ dx . Now substitute y=x+a2y=x+\dfrac{a}{2} to get 3a/2(2b+a)/21y2+ba24 dy\displaystyle\int_{3a/2}^{(2b+a)/2} \dfrac{1}{ y^2+ b-\dfrac{a^2}{4}} \ dy. Now using 1x2+a=tan1(xa)a+C\int \dfrac{1}{x^2+a} = \dfrac{\tan^{-1}\left(\dfrac{x}{\sqrt{a}}\right)}{\sqrt{a}}+C , and substituting limits , the answer IS 24ba2tan1(2(ba)(4ba2)2a2+6ab+4b)\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.

Nihar Mahajan - 3 years, 7 months ago

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PROBLEM 2:

01ln(x+a)x+a(x+b)x+b(x+a)(x+b)ln(x+a)ln(x+b) dx.\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.

Nihar Mahajan - 3 years, 7 months ago

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SOLUTION TO PROBLEM 2:

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

ln(x+a)x+a(x+b)x+b(x+a)(x+b)ln(x+a)ln(x+b)=1(x+b)ln(x+b)1(x+a)ln(x+a)\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 dx(x+k)ln(x+k)\displaystyle \int \dfrac{dx}{(x+k)\ln(x+k)}

ln(x+k)=t\ln(x+k)=t

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

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

Using this and putting limits we get the answer as :

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

Samarth Agarwal - 3 years, 7 months ago

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PROBLEM 3:

2x1x22x+10 dx \displaystyle \int \dfrac{2x-1}{x^2-2x+10} \ dx

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

Samarth Agarwal - 3 years, 7 months ago

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SOLUTION TO PROBLEM 3:

2(x1)+1x22x+10 dx\displaystyle \int \dfrac{2(x-1)+1}{x^2-2x+10} \ dx

=2(x1)x22x+10+1(x1)2+9 dx=\displaystyle \int \dfrac{2(x-1)}{x^2-2x+10}+\dfrac{1}{(x-1)^2+9} \ dx

=ln(x22x+10)+13(tan1(x13))+C=\ln(x^2-2x+10)+\dfrac 13(\tan^{-1}(\frac{x-1}{3}))+C

Rishabh Jain - 3 years, 7 months ago

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Problem 5:

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

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

This problem was solved by Vighnesh Shenoy

Samarth Agarwal - 3 years, 7 months ago

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Let us consider f(x) is a constant c.
We get,
c2=2c c^{2} = 2c
c=2 c= 2 or c=0 c = 0
Both of these do not satisfy,
f(10)=1001 f(10) = 1001
Now,
f(x)=f(1x)f(1x)1 f(x) = \dfrac{f\left(\dfrac{1}{x}\right)}{f\left(\dfrac{1}{x}\right) - 1}
(f(x)1)(f(1x)1)=1 \therefore \left( f(x)-1 \right) \cdot \left ( f\left(\dfrac{1}{x}\right) - 1 \right) = 1
Let,
f(x)=g(x)+1f(1x)=g(1x)+1 f(x) = g(x) + 1 \rightarrow f\left(\dfrac{1}{x}\right) = g\left(\dfrac{1}{x}\right) + 1
g(x)g(1x)=1 \therefore g(x) \cdot g\left(\dfrac{1}{x}\right) = 1
g(x) g(x) is a polynomial of the type ±xn \pm x^{n}
f(x)=1±xn \therefore f(x) = 1 \pm x^{n}
Substitute x = 10,
1001=1±10n 1001 = 1 \pm 10^{n}
±10n=1000 \pm 10^{n} = 1000
10n 10^{n} can not be negative.
10n=1000n=3 \therefore 10^{n} = 1000 \rightarrow n = 3
f(x)=x3+1 f(x) = x^{3} + 1
f(5)=125+1=126 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.

A Former Brilliant Member - 3 years, 7 months ago

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PROBLEM 10:

If,

S=cos3(x)12cos5(x)13cos7(x)14cos9(x)...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 π25π2Sdx\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.

Akshay Yadav - 3 years, 7 months ago

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Area under the graph of cos2n1(x)\cos^{2n-1} (x) from k to 2π+kk \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(1x2)=n=1x2nn\ln(1-x^2)=-\sum_{n=1}^{\infty} \frac{x^{2n}}{n}

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

cos(x)ln(sin2(x))dx\displaystyle \int \cos(x)\ln(\sin^2 (x))\mathrm{d}x

Vishnu Bhagyanath - 3 years, 7 months ago

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Is the answer 0 @Akshay Yadav

Samarth Agarwal - 3 years, 7 months ago

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Problem 13

In=ππ(sinnx(1+πx)sinx)dxI_n=\int_{-\pi}^{\pi}\left(\dfrac{\sin nx }{(1+\pi^x)\sin x}\right)\mathrm{d}x n=0,1,2,....n=0,1,2,....

In+2In=k×100!,kZI_{n+2}-I_n=k\times 100!, k\in \mathbf{Z}

Find k!k!

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

Rishabh Jain - 3 years, 7 months ago

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SOLUTION TO PROBLEM 13:

Note that In+2In=ππsin(nx+2x)sinnx(1+πx)sinx dx=ππ2sinxcos((n+1)x)(1+πx)sinx dx=ππ2cos((n+1)x)(1+πx) dxI_{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 aaE(x)1+πx dx=0aE(x) dx\displaystyle\int_{-a}^{a} \dfrac{E(x)}{1+\pi^x} \ dx = \displaystyle\int_{0}^{a} E(x) \ dx :

0π2cos((n+1)x)=2sin(n+1)n+10π=0\int_{0}^{\pi} 2 \cos ((n+1)x) = \dfrac{2\sin(n+1)}{n+1} \bigg|_{0}^{\pi} = 0

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

Nihar Mahajan - 3 years, 7 months ago

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In=ππsin(nx)dx(1+πx)sin(x)=0πsin(nx)dxsin(x) I_{n} = \displaystyle \int_{-\pi}^{\pi} \dfrac{\sin(nx)dx}{(1+\pi^{x})\sin(x)} = \int_{0}^{\pi} \dfrac{\sin(nx)dx}{\sin(x)}

In+2In=0πsin((n+2)x)sin(nx)dxsinx I_{n+2} - I_{n} = \displaystyle \int_{0}^{\pi} \dfrac{\sin((n+2)x) - \sin(nx) dx }{\sin x }
In+2In=0π2sin(x)cos((n+1)x)dxsinx=20πcos((n+1)x)dx=0 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 k = 0, k! = 1

A Former Brilliant Member - 3 years, 7 months ago

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PROBLEM 7:

Find all real numbers xx such that 0xt2sin(xt) dt=x2\displaystyle\int_0^x t^2\sin (x-t)\ dt=x^2

This problem has been solved by Akshay Yadav.

Nihar Mahajan - 3 years, 7 months ago

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SOLUTION TO PROBLEM 7:

0xt2sin(xt)dt=x2\int_0^{x} t^2 \sin(x-t) \mathrm{d}t=x^2

0xt2[sin(x)cos(t)sin(t)cos(x)]dt=x2\int_0^{x} t^2[ \sin(x)\cos(t)-\sin(t)\cos(x)] \mathrm{d}t=x^2

0xt2sin(x)cos(t)dt0xt2sin(t)cos(x)dt=x2\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)0xt2cos(t)dtcos(x)0xt2sin(t)dt=x2\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)[x2sin(x)2sin(x)+2xcos(x)]cos(x)[x2cos(x)+2xsin(x)+2cos(x)2]=x2\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

x2+2cos(x)2=x2x^2+2\cos(x)-2=x^2

cos(x)=1\cos(x)=1

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

Akshay Yadav - 3 years, 7 months ago

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PROBLEM 11:

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

This problem was solved by Vighnesh Shenoy.

Vishnu Bhagyanath - 3 years, 7 months ago

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x=t2dx=2tdt x = t^{2} \rightarrow dx = 2tdt
I=20tetdt I = \displaystyle 2 \int_{0}^{\infty}te^{-t}dt
tetdt=et(t+1) \displaystyle \int te^{-t}dt = -e^{-t}(t+1)
I=2[et(t+1)]0=2(limxet(t+1)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.

A Former Brilliant Member - 3 years, 7 months ago

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Find the general solution to the differential equation,
dydx2ytan(x)=sin(2x) \dfrac{dy}{dx} - 2y\tan(x )= \sin(2x)

This problem was solved by Rishabh Cool.

A Former Brilliant Member - 3 years, 7 months ago

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Integration factor =e2tanxdx=cos2xe^{\int -2\tan x dx}=\cos^2 x.

ycos2x=2sinxcos3xdxy\cos^2 x=\int 2\sin x \cos^3 xdx Make substitution cosx=t\cos x=t to evaluate the integral such that sinxdx=dt-\sin x dx=dt ycos2x=cos4x2+C\boxed{y\cos ^2 x=-\dfrac{\cos^4 x}{2}+C}

Rishabh Jain - 3 years, 7 months ago

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PROBLEM 14:

Suppose a,ba,b are real numbers such that a+b=1a+b=1. Then prove that the minimum value of the integral 0π(asinx+bsin2x)2 dx \displaystyle\int_{0}^{\pi}(a\sin x+b\sin 2x)^{2}\ dx is π4\dfrac{\pi}{4} and it occurs at a=b=12a=b=\dfrac{1}{2}. (Use Calculus only).

This problem was solved by Vighnesh Shenoy.

Nihar Mahajan - 3 years, 7 months ago

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I=0π(asinx+bsin2x)2dx I = \displaystyle \int_{0}^{\pi} (a\sin x + b\sin 2x)^{2} dx

I=0π2(asinx+bsin2x)2+(asinxbsin2x)2dx I = \displaystyle \int_{0}^{\dfrac{\pi}{2}} (a\sin x + b \sin 2x )^{2} + (a \sin x - b \sin 2x)^{2} dx

I=20π2a2sin2x+b2sin22xdx \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, 02af(x)dx=0af(x)+f(2ax)dx \displaystyle \int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) + f(2a-x) dx
I=20π4a2dx+40π4b2sin22xdx 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=2a2π4+4b2π8=πa2+b22=π×(a2+(1a)2)2 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,
dIda=π×2a2(1a)2=0a=12=b \dfrac{dI}{da} = \pi \times \dfrac{2a -2(1-a)}{2} = 0 \rightarrow a = \dfrac{1}{2} = b
Differentiate again with respect to a,
d2Ida2=π×2>0 \dfrac{d^{2}I}{da^{2}} =\pi \times 2 > 0
Thus the value is minimum, and occurs at a=b=12 a = b = \dfrac{1}{2}
Imin=π2×((12)2)×2=π4 I_{min} = \dfrac{\pi}{2} \times \left( \left(\dfrac{1}{2}\right)^{2} \right) \times 2 = \dfrac{\pi}{4}

A Former Brilliant Member - 3 years, 7 months ago

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I have few doubts, doubt killer:

1) How did you get second step from first?

2) How is 02af(x)dx=0af(x)+f(2ax)dx \displaystyle \int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) + f(2a-x) dx ?

BTW Nice solution , Post next problem vighu.

Nihar Mahajan - 3 years, 7 months ago

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@Nihar Mahajan In the first step I used the same property.
Proof :
I=02af(x)dx=0af(x)dx+a2af(x)dx I = \displaystyle \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)dx + \int_{a}^{2a}f(x)dx
In the second integral,
2ax=t,dx=dt 2a - x = t, \rightarrow dx = - dt
a2af(x)dx=a0f(2at)dt=0af(2at)dt=02af(2ax)dx \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
I=0af(x)+f(2ax)dx \therefore I = \displaystyle \int_{0}^{a}f(x) + f(2a-x)dx

A Former Brilliant Member - 3 years, 7 months ago

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@A Former Brilliant Member Thanks!

Nihar Mahajan - 3 years, 7 months ago

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PROBLEM 6:

Find the closed form of the indefinite integral,
dx1cotx \displaystyle \int \dfrac{dx}{1-\cot x}

This problem has been solved by Nihar Mahajan

A Former Brilliant Member - 3 years, 7 months ago

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

dx1cotx=122sinxsinxcosx dx=12sinx+cosx+sinxcosxsinxcosx dx=12[1 dx+sinx+cosxsinxcosx dx] \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=sinxcosxdu=cosx+sinx dxu=\sin x - \cos x \Rightarrow du = \cos x+\sin x \ dx , the above integral changes to:

12[x1u du]=xln(cosxsinx)2+C\dfrac{1}{2}\left[x-\int \dfrac{1}{u} \ du\right] = \dfrac{x-\ln(|\cos x - \sin x|)}{2} + C

Nihar Mahajan - 3 years, 7 months ago

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PROBLEM 8:

Find the length of an arc of curve

y=a2(exa+exa)y=\frac{a}{2}\left(e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right)

between x=0x=0 and x=ax=a.

Answer in terms of aa and ee in simplest form.

This problem has been solved by Nihar Mahajan.

Akshay Yadav - 3 years, 7 months ago

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SOLUTION TO PROBLEM 8:

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

1+(dydx)2=1+sinh2(xa)=cosh(xa)\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=0x=0 to x=ax=a is given by the integral: 0acosh(xa) dx\displaystyle\int_{0}^{a} \cosh\left(\dfrac{x}{a}\right) \ dx

Substitute y=xadx=a dyy=\dfrac{x}{a} \Rightarrow dx=a \ dy and changing limits , the integral becomes a[sinh(y)]01=asinh(1)=a(e21)2ea\left[\sinh(y)\right]\bigg|_{0}^{1} = a\sinh(1) = \dfrac{a(e^2-1)}{2e}

Nihar Mahajan - 3 years, 7 months ago

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PROBLEM 9:

Find the minimum area of the region bounded by the curve y=a3x2a4xy=a^3x^2-a^4x and the line y=xy=x where a>0a>0 .

This problem has been solved by Akshay Yadav.

Nihar Mahajan - 3 years, 7 months ago

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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 aa.

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 aa.

Differentiate it and find the global minima.

Akshay Yadav - 3 years, 7 months ago

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PROBLEM 15:

Evaluate the following integral,
0π24asin2x+6b2cos2x(a2sin2x+b3cos2x)2dx \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.

A Former Brilliant Member - 3 years, 7 months ago

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I2=0π22asin2x(a2sin2x+b3cos2x)2dx+0π23b2cos2x(a2sin2x+b3cos2x)2dx\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}dxI1(a)=0π2dxa2sin2x+b3cos2xI_1(a)=\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{a^2\sin^2x+b^3\cos^2x}I1(a)=0π2dx(a2sin2x+b3cos2x)2×2asin2x\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 I2(b)I_2(b),and hence I2(b)=0π2dx(a2sin2x+b3cos2x)2×3b2cos2xI_2'(b)=\int_0^{\frac{\pi}{2}}\dfrac{-dx}{(a^2\sin^2x+b^3\cos^2x)^2}\times 3b^2\cos^2x,hence,I1(a)+I2(b)=I2()I_1'(a)+I_2'(b)=-\dfrac{I}{2}(*),evaluating I1(a)I_1(a) and I2(b)I_2(b) using tan\tan and sec\sec technique and using the fact that dxx2+a2=1atan1x\int\dfrac{dx}{x^2+a^2}=\dfrac{1}{a}\tan^{-1}{x},we finally get that,I1(a)=1a2b32π2and I2(b)=321ab52π2I_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=π3a+2b2a2b52I=\pi\dfrac{3a+2b}{2a^2b^{\frac{5}{2}}},it is done!

Adarsh Kumar - 3 years, 7 months ago

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Problem 17:

f(x)=x12ax36a2f(x) = \sqrt{x - \sqrt{12ax-36a^{2}}}

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

This problem was solved by Nihar Mahajan.

Harsh Shrivastava - 3 years, 7 months ago

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SOLUTION TO PROBLEM 17:

Note that f(x)=(x3a)2+(3a)223ax9a2=(x3a3a)2=x3a3af(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, x3a3a=2(x3a)3/233ax+C\displaystyle\int \sqrt{x-3a}-\sqrt{3a} = \dfrac{2(x-3a)^{3/2}}{3} - \sqrt{3a}x+C .

Nihar Mahajan - 3 years, 7 months ago

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PROBLEM 19:

Find the quadratic mean and the arithmetical mean of the function y=A1sin(x)+A3sin(3x)y=A_1\sin (x)+A_3\sin (3x)

Subsequently find the minimum value of quadratic mean as A1,A3RA_1,A_3 \in \mathbb{R}.

This question was solved by Hummus A.

Akshay Yadav - 3 years, 7 months ago

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here's my solution

quadratic mean: first we square the function,we get

A12sin2x+2A1A3sin(x)sin(3x)+A32sin2(3x){ 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

1π0πA12sin2x+2A1A3sin(x)sin(3x)+A32sin2(3x)dx=1π(A120πsin2xdx+2A1A30πsin(x)sin(3x)dx+A320πsin2(3x)dx)=A122+A322\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 A1=A3=0{ 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

π/23π/2sinx+sin3xdx2π=02π=0\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 :)

Hamza A - 3 years, 7 months ago

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PROBLEM 20

prove that

abxdx(xa)(bx)=π(a+b)2\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.

Hamza A - 3 years, 7 months ago

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Solution to problem 20:abxdx(xa)(bx)\int_a^b\dfrac{x dx}{\sqrt{(x-a)(b-x)}}=ab(a+bx)dx(xa)(bx)=\int_a^b\dfrac{(a+b-x)dx}{\sqrt{(x-a)(b-x)}}2I=(a+b)ab(a+b)dx(xa)(bx)\Longrightarrow 2I=(a+b)\int_a^b(a+b)\dfrac{dx}{\sqrt{(x-a)(b-x)}}.Now substituting z=a+b2xz=\dfrac{a+b}{2}-x,and changing the limits accordingly,2I=(a+b)ba2ab2dz(ba2z)(ba2+z)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)}}dxa2x2=sin1xa\int \dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\dfrac{x}{a},doing the calculations we get,2I=(a+b)(π)I=(a+b)π22I=-(a+b)(-\pi)\\ \Longrightarrow I=\dfrac{(a+b)\pi}{2}.It is done!

Adarsh Kumar - 3 years, 7 months ago

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WOW!, 4 people at the same time

Hamza A - 3 years, 7 months ago

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

Samarth Agarwal - 3 years, 7 months ago

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You post the next question ;)

Nihar Mahajan - 3 years, 7 months ago

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First simplifying ab1(xa)(bx)dx\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.

ab1xbx2ab+axdx\displaystyle \int_a^b \frac{1}{\sqrt{xb-x^2-ab+ax}} \mathrm{d}x

ab1(xa+b2)2+(ab2)2dx\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(2xaba+bab)ab\arcsin \left(\frac{2x}{a-b}-\frac{a+b}{a-b}\right)|_a^b

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

Akshay Yadav - 3 years, 7 months ago

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SOLUTION TO PROBLEM 20:

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

I=abx(xa)(bx) dx=aba+bx(xa)(bx) dxI=\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=aba+b(xa)(bx) dx2I= \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=xabay=\dfrac{x-a}{b-a} and note that 1y=bxba1-y = \dfrac{b-x}{b-a} and (ba)dy=dx(b-a)dy=dx and the integral changes to: I=a+b2011y(1y) dyI=\dfrac{a+b}{2} \displaystyle\int_{0}^{1} \dfrac{1}{\sqrt{y(1-y)}} \ dy which is simply the antiderivative for arcsin\arcsin. So we have the integral evaluated as π(a+b)2\dfrac{\pi(a+b)}{2} .

Nihar Mahajan - 3 years, 7 months ago

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

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

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

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

=aba+b2(dx(ba)22(x((a+b)2))2)=\int_a^b\dfrac{a+b}{2}(\dfrac{dx}{ \sqrt {\dfrac{(b-a)^2}{2}- (x-(\dfrac{(a+b)}{2}))^2} })

Now using dxa2x2=sin1xa\int \dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1} \dfrac xa.

I=(a+b)2(sin1(2x(a+b)ba))abI=\dfrac{(a+b)}{2}(\sin^{-1}(\dfrac{2x-(a+b)}{b-a}))|_a^b

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

Rishabh Jain - 3 years, 7 months ago

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Problem 21 by Adarsh Kumar:[(x+a)3(x+b)5]14dx=?\int[(x+a)^{-3}(x+b)^{-5}]^{\frac{1}{4}}dx=?

This problem was first solved by Harsh Shrivastva.

Akshay Yadav - 3 years, 7 months ago

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Solution 21 by Harsh Shrivastava:

Akshay Yadav - 3 years, 7 months ago

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Problem 22

0πxlnsinx dx0πlnsinx dx=π/2\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.

Harsh Shrivastava - 3 years, 7 months ago

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I=0πxlnsinx dxI10πlnsinx dxI2I=\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}}}

I1=0πxlnsinx dxI_1=\int_{0}^{\pi} x\ln \sin x \ dx

=0π(πx)lnsin(πx) dx=\int_0^{\pi} (\pi-x)\ln \sin (\pi-x) \ dx

=0ππlnsinx dxI1=\int_{0}^{\pi} \pi\ln \sin x \ dx-I_1

    I1=π20πlnsinx dx=π2I2\implies I_1=\dfrac{\pi}{2}\int_{0}^{\pi} \ln \sin x \ dx=\dfrac{\pi}{2}I_2

And thus :

I1I2=π2=I\dfrac{I_1}{I_2}=\dfrac{\pi}{2}=I

Hence proved...

Rishabh Jain - 3 years, 7 months ago

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Post the next question please.

Adarsh Kumar - 3 years, 7 months ago

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PROBLEM 16:

dx(5+4cosx)2=?\int\dfrac{dx}{(5+4\cos x)^2}=?