Brilliant Sub Junior Calculus Contest (Season-1)

Solution to problem 4:

First we start with the following inequality:

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

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 $4\sqrt3-4-\dfrac{\pi}{3}$

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.

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 :

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

SOLUTION TO PROBLEM 3:

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

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

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

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.

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,

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

Is the answer 0 @Akshay Yadav

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

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

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

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,

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

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

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

$\cos(x)=1$

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

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

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

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 $-\sin x dx=dt$ $\boxed{y\cos ^2 x=-\dfrac{\cos^4 x}{2}+C}$

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

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

$\therefore I = \displaystyle 2\int_{0}^{\dfrac{\pi}{2}} a^{2} \sin^{2}x + b^{2} \sin^{2}2x dx$
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 $a = b = \dfrac{1}{2}$
$I_{min} = \dfrac{\pi}{2} \times \left( \left(\dfrac{1}{2}\right)^{2} \right) \times 2 = \dfrac{\pi}{4}$

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:

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

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 $a\left[\sinh(y)\right]\bigg|_{0}^{1} = a\sinh(1) = \dfrac{a(e^2-1)}{2e}$

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.

$\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!

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

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 :)

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!

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.

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

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

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

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

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 $\dfrac{\pi(a+b)}{2}$ .

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

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

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

Solution 21 by Harsh Shrivastava:

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

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

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

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

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

And thus :

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

Hence proved...

Substituting $\cos x = \frac{1-t^{2}}{1+t^{2}}$, where $t = \tan(x/2)$,the integral transforms into

$I = \displaystyle 2\int \dfrac{1+t^{2}}{(9+t^{2})^2}dt$

$I= \displaystyle 2\int\dfrac{1}{(9+t^{2})^2} dt-2\int \dfrac{t^2}{(9+t^{2})^2}dt$

In both the integrals,substitute $t = 3\tan p$,

Both the integrals will now be trivial to evaluate.

Sorry for this concise solution without any elaboration 'coz I am not in my home and using latex on mobile is very(!) cumbersome+tedious.

I will improve this solution when I'll reach home.

The integral given can be written as,

$\displaystyle \int_0^{1} -\frac{\cos((\beta+\alpha)x)-\cos((\beta-\alpha)x)}{2}\mathrm{d}x$

$\displaystyle \int_0^{1} -\frac{\cos((\beta+\alpha)x)}{2}\mathrm{d}x -\displaystyle \int_0^{1} -\frac{\cos((\beta-\alpha)x)}{2}\mathrm{d}x$

Applying linearity,

$-\frac{1}{2} \displaystyle \int_0^{1} \cos((\beta+\alpha)x)\mathrm{d}x+\frac{1}{2} \displaystyle \int_0^{1} \cos((\beta-\alpha)x)\mathrm{d}x$

Solving each integral we get,

$\dfrac{\alpha\cos(\alpha)\sin(\beta)-\sin(\alpha)\beta\cos(\beta)}{\beta^2-\alpha^2}$

$\dfrac{\cos(\alpha)\cos(\beta)(\alpha\tan(\beta)-\tan(\alpha)\beta)}{\beta^2-\alpha^2}$

Now,

$\tan(\alpha)=2\alpha$ and $\tan(\beta)=2\beta$

So,

$\dfrac{\cos(\alpha)\cos(\beta)(\alpha(2\beta)-(2\alpha)\beta)}{\beta^2-\alpha^2}$

Hence its equal to $\boxed{0}$.

Write it as : $\int (\dfrac{y-1}{y})^{\frac{-5}{4}}\dfrac{dy}{y^2}$ Substitute $\dfrac{y-1}{y}=t$ such that $\dfrac{dy}{y^2}=dt$.