Brilliant Sub Junior Calculus Contest (Season-1) Note-2

Solution without using Beta function: $\dfrac{5050\overbrace{\displaystyle \int_0^1(1-x^{50})^{100}\,\mathrm{d}x}^{\color{#D61F06}{I_1}}}{\underbrace{\displaystyle \int_0^1(1-x^{50})^{101}\,\mathrm{d}x}_{\color{#D61F06}{I_2}}}$

Applying IBP on $I_2$ $I_2=x(1-x^{50})^{101}|_0^1+5050\displaystyle \int_0^1 x^{50}(1-x^{50})^{100}\,\mathrm{d}x$ $=0+5050\displaystyle \int_0^1(\color{#20A900}{1}-(\color{#20A900}{1}- x^{50}))(1-x^{50})^{100}\,\mathrm{d}x$

$=5050\left(\displaystyle \int_0^1(1-x^{50})^{100}\,\mathrm{d}x-\displaystyle \int_0^1(1-x^{50})^{101}\,\mathrm{d}x\right)$

$\implies I_2=5050(I_1-I_2)$

Rearranging we get:

$\dfrac{5050I_1}{I_2}=\boxed{5051}$

Problem 25 : Let us first evaluate the numerator & denote it as $I_1$

Note : $\mathbf{Beta}$ Function is used which is as follows just to recall :

$\large \int_{0}^{1} x^{m-1}(1-x)^{n-1}=\beta(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\frac{(m-1)!(n-1)!}{(m+n-1)!}$ with m,n > 0.

Let $1-x^{50}=t$

$-50x^{49}dx=dt$

$x^{49} = (1-t)^{\frac{49}{50}}$ & $dx=\frac{dt}{-50(1-t)^{\frac{49}{50}}}$

The integral is changed to :

$I_1=-\frac{1}{50}\int_{1}^{0} t^{100}(1-t)^{-\frac{49}{50}}dt$

$50I_1=\int_{0}^{1} t^{101-1} (1-t)^{\frac{1}{50}-1}dt$

$50I_1=\beta(101,\frac{1}{50}=\frac{\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma(101+\frac{1}{50})}$

Similarly the denominator just differs in case of the exponent of t, denote the denominator by $I_2$ so we have ,

$50I_2=\beta(102,\frac{1}{50})=\frac{\Gamma(102)\Gamma(\frac{1}{50})}{\Gamma(102+\frac{1}{50})}$

$50I_2=\frac{101\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma([1]+[101+\frac{1}{50}])}$

Since we have $\Gamma(n+1)=n\Gamma(n)$ so ,

$50I_2= \frac{101}{101+\frac{1}{50}}\frac{\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma(101+\frac{1}{50})}$

$50I_2=\frac{101.50}{5051}.50I_1$

$5051=\frac{5050 . I_1}{I_2}$

Therefore ,

$\large \frac{5050.\int_{0}^{1}(1-x^{50})^{100}dx}{\int_{0}^{1}(1-x^{50})^{101}dx} = 5051$ $\large \boxed{Proved}$

$\large I=\int_{0}^{\pi} \frac{xdx}{1+\cos(\alpha)\sin(x)}$

Using $\int_a^b f(x)dx=\int_a^b f(a+b-x) dx$ and adding both integrals and simplifying gives:

$I=\dfrac{\pi}{2}\left(\int_0^{\pi}\dfrac{dx}{1+(\cos \alpha)( \sin x)}\right)$ Using $\sin x=\dfrac{2\tan^2 \frac x2}{1+\tan^2\frac x2}$ and substituting $\tan^2 \frac x2=t$ such that $\dfrac{(1+\tan^2 \frac x2 )dx}{2}=dt$ .

$I={\pi}\int_{0}^{\infty}\left(\dfrac{dt}{1+t^2+2\cos \alpha t}\right)$

$I={\pi}\int_{0}^{\infty}\left(\dfrac{dt}{(t+\cos \alpha )^2+\sin^2 \alpha }\right)$

Using $\int \dfrac{dx}{x^2+a^2}=\frac 1a(\tan^{-1} \frac xa)$ .

$=\dfrac{\pi}{\sin \alpha}\left(\tan^{-1}\left(\dfrac{t+\cos \alpha }{\sin \alpha}\right)\right)|_0^{\infty}$

$=\dfrac{\pi}{\sin \alpha}(\dfrac{\pi}{2}-\underbrace{\tan^{-1} \left(\cot \alpha\right)}_{\dfrac{\pi}{2}-\alpha})$

$\large \boxed{=\dfrac{\pi\alpha}{\sin \alpha}}$

Solution to Problem 24

Substitute $\ln \tan x=t$ such that $\dfrac{dx}{\sin 2x}=\dfrac{dt}{2}$. Integral transforms to: $\int \dfrac{dt}{2 t^3}$ $=\dfrac{-1}{ 4t^2}=\dfrac{-1}{4\ln^2 \tan x}+C$

Let us consider a polynomial $f(x) = a + bx + cx^2 + dx^3 + ex^4 + fx^5 + gx^6 + .......$.

$f'(x) = b + 2cx + 3dx^2 + 4ex^3 + 5fx^4 +6gx^5 + 7hx^6 + .......$.

$f''(x) = 2c + 6dx + 12ex^2 + 20fx^3 + 30gx^4 + 42hx^5 + .......$.

Since , $f(0) = 2 \implies a=2$ and $f'(0) = 3 \implies b=3$.

On comparing coefficient of $x^n$ in equation $f(x) = f''(x)$

$c=1 , d= \frac{1}{2} , e =\frac{1}{12} , f=\frac{1}{40} , g=\frac{1}{360} , h =\frac{1}{1680}$

We have , $f(4) = 2 + 12 + 16 +\frac{4^3}{2}+\frac{4^4}{12} +\frac{4^5}{40} + \frac{4^6}{360}$.

Or $f(4) = \displaystyle \sum_{n=1}^\infty \frac{ 4^n ( (-1)^n + 5 ) }{ 8(n-1)!} = \displaystyle \sum_{n=1}^\infty \frac{ 4^n (-1)^n }{ 8(n-1)!} + \frac{5}{2} \displaystyle \sum_{n=1}^\infty \frac{ 4^{(n-1)} }{ (n-1)!} = \frac{1}{8} \frac{-4}{e^4} + \frac{5}{2} e^4 = \boxed{ \frac{ 5e^8 - 1}{2e^4} }$

The problem is of a book named "Integral calculus by Amit M Agaarwal" The solution he gave I am just posting it here.Now I am finding the mistake in his or sachin's solution.

$f''(x)=f(x)\implies 2f'(x)f''(x)=2f(x)f'(x)\implies d({f'(x)}^2) = d({f(x)}^2)$

$\implies {f'(x)}^2 = {f(x)}^2 + C$

Utilising the conditions for $f(0) = 2 , f'(0) = 3$ we get $C=5$

We have $f'(x) = \sqrt{5+{f(x)}^2}$

$\int dx = \int \frac{d({f(x)})}{\sqrt{5+{f(x)}^2}}$

$x + C_1 = log|f(x) + \sqrt{5+{f(x)}^2}|$

again by the conditions putting x=0 we get ,

$C_1=log5$

So, $x=log\frac{|f(x) + \sqrt{5+{f(x)}^2}|}{5}$ ..................................... (1)

Observe that , $log\frac{|f(x) + \sqrt{5+{f(x)}^2}|}{5} = log\frac{5}{|\sqrt{5+{f(x)}^2}-f(x)|}$

From (1) ,

$5e^x = f(x) + \sqrt{5+{f(x)}^2}$ & similarly $5e^{-x}=\sqrt{5+{f(x)}^2}-f(x)$

Extracting f(x) we get ,

$2f(x) = 5(e^x-e^{-x})\implies f(x) = \frac{5(e^x-e^{-x})}{2}$

$\large\boxed{ f(4) = \frac{5(e^4-e^{-4})}{2} = \frac{5(e^8-1)}{2e^4}}$

The problem is incorrect. If a polynomial $f(x)$ satisfies $f''(x) = f(x)$, then, even without solving the differential equation, the number of roots of $f(x)$ equals the number of roots of $f''(x)$, which is a contradiction. So it must be stated that $f$ is a function, not a polynomial function.

Solution to Problem 28:

There is a quick technique to solve the summations of this kind..... Slash of the first term and write down the other terms as it is in deno. and multiply by the number of remaining terms in the denominator (here 'p') now multiply this result with (-1) and add a constant C so the result till now is : $\\ S=C-\dfrac{1}{p(x+1)(x+2)\cdots(x+p)} \\$

When $x=1, \ S=\dfrac{1}{(p+1)!} \ also \ S=C-\dfrac{1}{p(p+1)!} \\$ This gives $C=\dfrac{1}{p.p!}$

In both the integrals $\dfrac{1}{p^2(p-1)!}-S$ is equivalent to $C-(C-\dfrac{1}{p(x+1)(x+2)\cdots(x+p)}) \\$ which is equivalent to $\dfrac{x!}{p(x+p)!}$

Now the integral in numerator turns into

$\displaystyle \int_0^{2016} dx$ and that in denominator turns into

$\displaystyle \int_0^{2} (x)(x-1)dx$

Thus, $\varphi=3024$ and answer is 4.

$I = \displaystyle \int \frac{\mathrm{d}x}{x^{11}\sqrt{1+x^4}}=$

Let $\frac{1}{x^4} = t$

So , $I = - \frac{1}{4} \displaystyle \int \dfrac{t^2 \mathrm{d}t}{\sqrt{1+t}}$

Adding and subtracting 1 in numerator , And integrating ,

We have $I = - \frac{1}{4} {[ \frac{2}{15} (t+1)^{3/2} ( 3t -7) + 2\sqrt{t+1} ]} + C$

After putting the value of t in terms of x and simplifying , Now , $I = - \frac{ \sqrt{x^4 + 1}( 8x^8 - 4x^4 + 3)}{30x^{10}} + C$

Applying LHopital:

$\lim_{x\to 0}\dfrac{8x\ln(1+x)}{x^2(x^4+4)}$

$=8\lim_{x\to 0}\dfrac{\ln(1+x)}{x(4)}$

$\text{Using} \lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1$

$=\boxed 2$

$I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{2}} f(\sin 2x)\sin x dx$
Using the property,
$\displaystyle \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)+f(2a-x)dx$
$I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x ) dx$
In $I_{2}$ ,
Use $\displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b} f(a+b-x)dx$
$\therefore I_{2} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)\cos\left( \dfrac{\pi}{4} - x\right) dx$
$\therefore I_{2} = \dfrac{1}{\sqrt{2}}\cdot \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x) dx = \dfrac{I_{1}}{\sqrt{2}}$
$\dfrac{I_{1}}{I_{2}} = \sqrt{2}$

$\tan x = t^{2}$
$dx = \dfrac{2t}{1+t^{4}}dt$
$I = \displaystyle \int \dfrac{2t^{2}}{1+t^{4}}dt$
$I = \displaystyle \int \dfrac{t^{2}-1}{1+t^{4}}dt + \int \dfrac{t^{2}+1}{1+t^{4}}dt$

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

$I = \displaystyle \int \dfrac{1-\frac{1}{t^{2}}}{\left(t+\frac{1}{t} \right)^{2} - 2}dt + \int \dfrac{1+\frac{1}{t^{2}}}{\left(t-\dfrac{1}{t}\right)^{2} + 2 }dt$
$I = \displaystyle \int \dfrac{du}{u^{2} - (\sqrt{2})^{2}} + \int \dfrac{dv}{v^{2}+(\sqrt{2})^{2}}$
$I =\dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{u-\sqrt{2}}{u+\sqrt{2}}\right) + \dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{v}{\sqrt{2}}\right)$

$I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)+ \dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{t-\frac{1}{t}}{\sqrt{2}}\right)$

$I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{\sqrt{\tan(x)} + \sqrt{\cot(x)}-\sqrt{2}}{\sqrt{\tan(x)} + \sqrt{\cot(x)}+\sqrt{2}}\right)+\dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{\sqrt{\tan(x)}-\sqrt{\cot(x)}}{\sqrt{2}}\right) + c$

@Rishabh Cool is this your original problem??? Btw Its really very nice

I don't have a good problem right now.... Can anybody else post the question... Thanx

My method was to take denominator as t and then apply a little partial fraction giving the same answer... :)

@Aditya Sharma you may post the next question

$\large I = \int \frac{tanx}{a+b(tan)^2x}\ = \int \frac{sin(x)cos(x)}{acos^2x+bsin^2x}$

$\text{Put } sin^2x=t$

$I = \frac{1}{2}\int \frac{sin(2x)}{a - sin^2x(a-b)} = \int \frac{dt}{2(a - t(a-b))} = \frac{1}{2(b-a)}ln|a-t(a-b)| + C$

$I = \displaystyle \int_{3}^{3e - \frac{1}{2}} g(x+1) dx = \int_{4}^{3e+\frac{1}{2}} g(x)dx = \int_{f(0)}^{f(1)}g(x)dx$
Substitute, $g(x) = t \rightarrow f(g(x)) = f(t)$
But since, $f(x)$ and $g(x)$ are inverses of each other, $f(g(x))= x$
$\therefore x = f(t) \rightarrow dx = f'(t)dt$
$\therefore I = \displaystyle \int_{0}^{1} tf'(t)dt$
Integrating by parts,
$I = \left[tf(t)\right]_{0}^{1} - \displaystyle \int_{0}^{1} f(t)dt$
$I = 3e + \dfrac{1}{2} - \displaystyle \int_{0}^{1} 3e^{t} + \dfrac{1}{1+t^{2}}dt = 3e + \dfrac{1}{2} - \left[ 3e^{t} + \tan^{1}(x) \right]_{0}^{1}$
$\therefore I = 3e + \dfrac{1}{2} -3e - \dfrac{\pi}{4} + 3 + 0 = \dfrac{7}{2} - \dfrac{\pi}{4} = \dfrac{14-\pi}{4}$

I had simplified the expression and then integrated it, after simplification you get-

$\displaystyle \int_0^{\pi} \left(x^2+\pi^2a^2+\frac{b^2}{\pi^2}\cos^2 (x)-2\pi ax+2ab\cos (x)-\frac{2b}{\pi}\cos (x) \right) dx$

From that you get the expression-

$\frac{3b^2+24b+6{\pi}^4a^2-6{\pi}^4a+2{\pi}^4}{6{\pi}}$

Let this equals to $y$ and then we use partial differentiation,

From that we get $a=\frac{1}{2}$ and $b=-4$.

Place the values back and then you'll get the required minimum value of the integral as,

$\boxed{\frac{\pi^3}{12}-\frac{8}{\pi}}$

It is a semi circle above x axis as y is always positive with radius 1 unit and center (2,0)

$\dfrac{f'(x)}{f(x)}=-1$

Integrate both sides to get: $f(x)=Ce^{-x}$ $\implies g(x)=f'(x)=-Ce^{-x}$ $\implies h(x)=2C^2e^{-2x}$

$\dfrac{h(10)}{h(5)}=\dfrac{e^{-20}}{e^{-10}}=e^{-10}$

$\implies h(10)=\boxed{11e^{-10}}$

Is it or not???

The given integral can be written as

$\displaystyle \int \dfrac{\cos x \ dx}{1+2 \sin x \cos x}$

$\displaystyle = \int \dfrac{\cos x \ dx}{(\sin x + \cos x )^2}$

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

$\displaystyle = \dfrac{1}{2} \int \dfrac{dx}{\sin x + \cos x} + \dfrac{1}{2} \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2}$

$\displaystyle = \dfrac{1}{2\sqrt{2}} \int \csc \left(x + \frac{\pi}{4}\right) dx + \dfrac{1}{2} \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2}$

$\displaystyle = -\dfrac{1}{2\sqrt{2}} \ln \left| \csc\left( x +\frac{\pi}{4}\right) + \cot \left(x+\frac{\pi}{4}\right) \right|-\dfrac{1}{2(\sin x +\cos x)} + C$

Is the answer long?

@Samuel Jones Problem 40 on the new note please