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

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

Click here for First Note (if you want to visit it)

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.

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  • 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.

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

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**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, 6 months ago

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Problem 23 x21x32x42x2+1dx= ?\large\int\dfrac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx=~?

This problem has been solved by Harsh Shrivastava.

Rishabh Jain - 3 years, 6 months ago

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Harsh Shrivastava - 3 years, 6 months ago

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A similar problem like this came in JEE Mains 2016! I just saw the question paper and recalled this problem ;)

Nihar Mahajan - 3 years, 6 months ago

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@Nihar Mahajan Yeah.

Harsh Shrivastava - 3 years, 6 months ago

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Problem 25

Prove that:

505001(1x50)100dx01(1x50)101dx=5051\dfrac{5050\displaystyle \int_0^1(1-x^{50})^{100}\,\mathrm{d}x}{\displaystyle \int_0^1(1-x^{50})^{101}\,\mathrm{d}x}=\boxed{5051}

This problem has been solved by Aditya Sharma.

Rishabh Jain - 3 years, 6 months ago

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Solution without using Beta function: 505001(1x50)100dxI101(1x50)101dxI2\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 I2I_2 I2=x(1x50)10101+505001x50(1x50)100dxI_2=x(1-x^{50})^{101}|_0^1+5050\displaystyle \int_0^1 x^{50}(1-x^{50})^{100}\,\mathrm{d}x =0+505001(1(1x50))(1x50)100dx=0+5050\displaystyle \int_0^1(\color{#20A900}{1}-(\color{#20A900}{1}- x^{50}))(1-x^{50})^{100}\,\mathrm{d}x

=5050(01(1x50)100dx01(1x50)101dx)=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)

    I2=5050(I1I2)\implies I_2=5050(I_1-I_2)

Rearranging we get:

5050I1I2=5051\dfrac{5050I_1}{I_2}=\boxed{5051}

Rishabh Jain - 3 years, 6 months ago

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Both methods are same. The proof of Gamma functional equation uses IBP, so both methods are essentially the same.

Samuel Jones - 3 years, 6 months ago

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Problem 25 : Let us first evaluate the numerator & denote it as I1I_1

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

01xm1(1x)n1=β(m,n)=Γ(m)Γ(n)Γ(m+n)=(m1)!(n1)!(m+n1)!\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 1x50=t1-x^{50}=t

50x49dx=dt-50x^{49}dx=dt

x49=(1t)4950x^{49} = (1-t)^{\frac{49}{50}} & dx=dt50(1t)4950dx=\frac{dt}{-50(1-t)^{\frac{49}{50}}}

The integral is changed to :

I1=15010t100(1t)4950dtI_1=-\frac{1}{50}\int_{1}^{0} t^{100}(1-t)^{-\frac{49}{50}}dt

50I1=01t1011(1t)1501dt50I_1=\int_{0}^{1} t^{101-1} (1-t)^{\frac{1}{50}-1}dt

50I1=β(101,150=Γ(101)Γ(150)Γ(101+150)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 I2I_2 so we have ,

50I2=β(102,150)=Γ(102)Γ(150)Γ(102+150)50I_2=\beta(102,\frac{1}{50})=\frac{\Gamma(102)\Gamma(\frac{1}{50})}{\Gamma(102+\frac{1}{50})}

50I2=101Γ(101)Γ(150)Γ([1]+[101+150])50I_2=\frac{101\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma([1]+[101+\frac{1}{50}])}

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

50I2=101101+150Γ(101)Γ(150)Γ(101+150)50I_2= \frac{101}{101+\frac{1}{50}}\frac{\Gamma(101)\Gamma(\frac{1}{50})}{\Gamma(101+\frac{1}{50})}

50I2=101.505051.50I150I_2=\frac{101.50}{5051}.50I_1

5051=5050.I1I25051=\frac{5050 . I_1}{I_2}

Therefore ,

5050.01(1x50)100dx01(1x50)101dx=5051\large \frac{5050.\int_{0}^{1}(1-x^{50})^{100}dx}{\int_{0}^{1}(1-x^{50})^{101}dx} = 5051 Proved\large \boxed{Proved}

Aditya Narayan Sharma - 3 years, 6 months ago

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Problem\large \mathbf{Problem} 27\large \mathfrak{27}

Prove that : 0πxdx1+cos(α)sin(x)=απsin(α)\large \int_{0}^{\pi} \frac{xdx}{1+cos(\alpha)sin(x)} = \frac{\alpha\pi}{sin(\alpha)}

Aditya Narayan Sharma - 3 years, 6 months ago

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I=0πxdx1+cos(α)sin(x)\large I=\int_{0}^{\pi} \frac{xdx}{1+\cos(\alpha)\sin(x)}

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

I=π2(0πdx1+(cosα)(sinx))I=\dfrac{\pi}{2}\left(\int_0^{\pi}\dfrac{dx}{1+(\cos \alpha)( \sin x)}\right) Using sinx=2tan2x21+tan2x2\sin x=\dfrac{2\tan^2 \frac x2}{1+\tan^2\frac x2} and substituting tan2x2=t\tan^2 \frac x2=t such that (1+tan2x2)dx2=dt\dfrac{(1+\tan^2 \frac x2 )dx}{2}=dt .

I=π0(dt1+t2+2cosαt)I={\pi}\int_{0}^{\infty}\left(\dfrac{dt}{1+t^2+2\cos \alpha t}\right)

I=π0(dt(t+cosα)2+sin2α)I={\pi}\int_{0}^{\infty}\left(\dfrac{dt}{(t+\cos \alpha )^2+\sin^2 \alpha }\right)

Using dxx2+a2=1a(tan1xa)\int \dfrac{dx}{x^2+a^2}=\frac 1a(\tan^{-1} \frac xa) .

=πsinα(tan1(t+cosαsinα))0=\dfrac{\pi}{\sin \alpha}\left(\tan^{-1}\left(\dfrac{t+\cos \alpha }{\sin \alpha}\right)\right)|_0^{\infty}

=πsinα(π2tan1(cotα)π2α)=\dfrac{\pi}{\sin \alpha}(\dfrac{\pi}{2}-\underbrace{\tan^{-1} \left(\cot \alpha\right)}_{\dfrac{\pi}{2}-\alpha})

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

Rishabh Jain - 3 years, 6 months ago

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Problem 24:

Evaluate dx(ln3(tanx))(sin2x)\displaystyle \int \dfrac{dx}{(\ln^{3} (\tan x))(\sin 2x)}

This problem was solved by Rishabh Cool.

Harsh Shrivastava - 3 years, 6 months ago

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Solution to Problem 24

Substitute lntanx=t\ln \tan x=t such that dxsin2x=dt2\dfrac{dx}{\sin 2x}=\dfrac{dt}{2}. Integral transforms to: dt2t3\int \dfrac{dt}{2 t^3} =14t2=14ln2tanx+C=\dfrac{-1}{ 4t^2}=\dfrac{-1}{4\ln^2 \tan x}+C

Rishabh Jain - 3 years, 6 months ago

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Problem\mathbf{Problem} 26\mathbf{26}

Let f(x)f(x) be a function satisfying f(0)=2,f(0)=3,f(x)=f(x)f(0)=2,f'(0)=3,f''(x)=f(x) , then prove that :

f(4)=5(e81)2e4f(4) = \frac{5(e^8-1)}{2e^4}

Aditya Narayan Sharma - 3 years, 6 months ago

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Let us consider a polynomial f(x)=a+bx+cx2+dx3+ex4+fx5+gx6+....... f(x) = a + bx + cx^2 + dx^3 + ex^4 + fx^5 + gx^6 + ....... .

f(x)=b+2cx+3dx2+4ex3+5fx4+6gx5+7hx6+....... f'(x) = b + 2cx + 3dx^2 + 4ex^3 + 5fx^4 +6gx^5 + 7hx^6 + ....... .

f(x)=2c+6dx+12ex2+20fx3+30gx4+42hx5+....... f''(x) = 2c + 6dx + 12ex^2 + 20fx^3 + 30gx^4 + 42hx^5 + ....... .

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

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

c=1,d=12,e=112,f=140,g=1360,h=11680 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+432+4412+4540+46360 f(4) = 2 + 12 + 16 +\frac{4^3}{2}+\frac{4^4}{12} +\frac{4^5}{40} + \frac{4^6}{360} .

Or f(4)=n=14n((1)n+5)8(n1)!=n=14n(1)n8(n1)!+52n=14(n1)(n1)!=184e4+52e4=5e812e4 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} }

Sachin Vishwakarma - 3 years, 6 months ago

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

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

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

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

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

x+C1=logf(x)+5+f(x)2x + C_1 = log|f(x) + \sqrt{5+{f(x)}^2}|

again by the conditions putting x=0 we get ,

C1=log5C_1=log5

So, x=logf(x)+5+f(x)25x=log\frac{|f(x) + \sqrt{5+{f(x)}^2}|}{5} ..................................... (1)

Observe that , logf(x)+5+f(x)25=log55+f(x)2f(x)log\frac{|f(x) + \sqrt{5+{f(x)}^2}|}{5} = log\frac{5}{|\sqrt{5+{f(x)}^2}-f(x)|}

From (1) ,

5ex=f(x)+5+f(x)25e^x = f(x) + \sqrt{5+{f(x)}^2} & similarly 5ex=5+f(x)2f(x)5e^{-x}=\sqrt{5+{f(x)}^2}-f(x)

Extracting f(x) we get ,

2f(x)=5(exex)    f(x)=5(exex)22f(x) = 5(e^x-e^{-x})\implies f(x) = \frac{5(e^x-e^{-x})}{2}

f(4)=5(e4e4)2=5(e81)2e4\large\boxed{ f(4) = \frac{5(e^4-e^{-4})}{2} = \frac{5(e^8-1)}{2e^4}}

Aditya Narayan Sharma - 3 years, 6 months ago

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

Samuel Jones - 3 years, 6 months ago

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This is correct ^. In reality the function is exponential.

A Former Brilliant Member - 3 years, 6 months ago

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Problem 28 S=n=1x(1n(n+1)(n+2)(n+p))\large\mathfrak{S}=\sum_{n=1}^x\left(\dfrac{1}{n(n+1)(n+2)\cdots(n+p)}\right)

02016((1p2(p1)!S)p(x+p)!x!)dx02((1p2(p1)!S)p(x+p)!(x2)!)dx=φ\dfrac{\displaystyle\int_0^{2016}\left(\left(\frac1{p^2(p-1)!}-\mathfrak S\right)\dfrac{p(x+p)!}{x!}\right)dx}{\displaystyle\int_0^{2}\left(\left(\frac1{p^2(p-1)!}-\mathfrak S\right)\dfrac{p(x+p)!}{(x-2)!}\right)dx}= \color{#3D99F6}{\varphi}

Find 602φ3!\dfrac{\sqrt{60^2-\color{#3D99F6}{\varphi}}}{3!}.

**Original

Rishabh Jain - 3 years, 6 months ago

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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=C1p(x+1)(x+2)(x+p)\\ S=C-\dfrac{1}{p(x+1)(x+2)\cdots(x+p)} \\

When x=1, S=1(p+1)! also S=C1p(p+1)!x=1, \ S=\dfrac{1}{(p+1)!} \ also \ S=C-\dfrac{1}{p(p+1)!} \\ This gives C=1p.p!C=\dfrac{1}{p.p!}

In both the integrals 1p2(p1)!S\dfrac{1}{p^2(p-1)!}-S is equivalent to C(C1p(x+1)(x+2)(x+p))C-(C-\dfrac{1}{p(x+1)(x+2)\cdots(x+p)}) \\ which is equivalent to x!p(x+p)!\dfrac{x!}{p(x+p)!}

Now the integral in numerator turns into

02016dx\displaystyle \int_0^{2016} dx and that in denominator turns into

02(x)(x1)dx\displaystyle \int_0^{2} (x)(x-1)dx

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

Samarth Agarwal - 3 years, 6 months ago

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Problem 30

dxx111+x4= ?\large\displaystyle \int \dfrac{\mathrm{d}x}{x^{11}\sqrt{1+x^4}}=\ ?

Rishabh Jain - 3 years, 6 months ago

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I=dxx111+x4= I = \displaystyle \int \frac{\mathrm{d}x}{x^{11}\sqrt{1+x^4}}=

Let 1x4=t \frac{1}{x^4} = t

So , I=14t2dt1+t 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=14[215(t+1)3/2(3t7)+2t+1]+C 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=x4+1(8x84x4+3)30x10+C I = - \frac{ \sqrt{x^4 + 1}( 8x^8 - 4x^4 + 3)}{30x^{10}} + C

Sachin Vishwakarma - 3 years, 6 months ago

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Problem 29:

Prove that:

limx024x30xtln(1+t)t4+4dt=2\displaystyle \lim_{x\to 0} \dfrac{24}{x^3} \displaystyle \int_0^x \dfrac{t \ln(1+t)}{t^4+4} dt =2

Samarth Agarwal - 3 years, 6 months ago

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Applying LHopital:

limx08xln(1+x)x2(x4+4)\lim_{x\to 0}\dfrac{8x\ln(1+x)}{x^2(x^4+4)}

=8limx0ln(1+x)x(4)=8\lim_{x\to 0}\dfrac{\ln(1+x)}{x(4)}

Usinglimx0ln(1+x)x=1\text{Using} \lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1

=2=\boxed 2

Rishabh Jain - 3 years, 6 months ago

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Problem 31

If I1=0π2f(sin2x)sinxdx I_{1} = \displaystyle \int_0^{\frac{\pi}{2} } f( \sin 2x) \sin x \mathrm{d}x and I2=0π4f(cos2x)cosxdx I_{2} = \displaystyle \int_0^{\frac{\pi}{4} } f( \cos 2x) \cos x \mathrm{d}x , then find I1I2 \frac{I_{1}}{I_{2}} ?

Sachin Vishwakarma - 3 years, 6 months ago

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I1=0π2f(sin2x)sinxdx I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{2}} f(\sin 2x)\sin x dx
Using the property,
02af(x)dx=0af(x)+f(2ax)dx \displaystyle \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)+f(2a-x)dx
I1=0π4f(sin2x)(cosx+sinx)dx I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x ) dx
In I2 I_{2} ,
Use abf(x)dx=abf(a+bx)dx \displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b} f(a+b-x)dx
I2=0π4f(sin2x)cos(π4x)dx \therefore I_{2} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)\cos\left( \dfrac{\pi}{4} - x\right) dx
I2=120π4f(sin2x)(cosx+sinx)dx=I12 \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}}
I1I2=2 \dfrac{I_{1}}{I_{2}} = \sqrt{2}

A Former Brilliant Member - 3 years, 6 months ago

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Let 's define the term,
In=0arctan(2x1)+arctan(3x2)+arctan((n+1)xn)narctan(x)xdx I_{n} = \displaystyle \int_{0}^{\infty} \dfrac{\arctan\left(\frac{2x}{1}\right) + \arctan\left(\frac{3x}{2}\right) + \ldots \arctan\left(\frac{(n+1)x}{n}\right) - n\arctan(x)}{x} dx for integer n.

Find I20I6 I_{20} - I_{6}

A Former Brilliant Member - 3 years, 6 months ago

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PROBLEM 33 tanx dx\displaystyle \int \sqrt{\tan x} \ dx

The answer is very long so be patient.!

Samarth Agarwal - 3 years, 6 months ago

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tanx=t2 \tan x = t^{2}
dx=2t1+t4dt dx = \dfrac{2t}{1+t^{4}}dt
I=2t21+t4dt I = \displaystyle \int \dfrac{2t^{2}}{1+t^{4}}dt
I=t211+t4dt+t2+11+t4dt I = \displaystyle \int \dfrac{t^{2}-1}{1+t^{4}}dt + \int \dfrac{t^{2}+1}{1+t^{4}}dt

I=11t2t2+1t2dt+1+1t2t2+1t2dt 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=11t2(t+1t)22dt+1+1t2(t1t)2+2dt 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=duu2(2)2+dvv2+(2)2 I = \displaystyle \int \dfrac{du}{u^{2} - (\sqrt{2})^{2}} + \int \dfrac{dv}{v^{2}+(\sqrt{2})^{2}}
I=122ln(u2u+2)+12arctan(v2) 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=122ln(t+1t2t+1t+2)+12arctan(t1t2) 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=122ln(tan(x)+cot(x)2tan(x)+cot(x)+2)+12arctan(tan(x)cot(x)2)+c 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

A Former Brilliant Member - 3 years, 6 months ago

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Will you post next problem or shall we post? You often stall the contest.

(Sorry to be rude)

Harsh Shrivastava - 3 years, 6 months ago

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@Harsh Shrivastava You post. I have been trying to come up with a good problem, but I haven't been getting one.

A Former Brilliant Member - 3 years, 6 months ago

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@A Former Brilliant Member Okay according to rules @Samarth Agarwal should post the next problem.

Harsh Shrivastava - 3 years, 6 months ago

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Very nice..... Pls post the next question

Samarth Agarwal - 3 years, 6 months ago

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Problem 37

Let f(x)f(x) be a differentiable function such that df(x)dx=f(x)+02f(x)dx \displaystyle \dfrac{d f(x) }{dx} = f(x) + \int_0^2 f(x) \, dx and f(0)=4e23f(0) = \dfrac{4-e^2}3 , then prove: f(2)=2e2+13f(2)=\dfrac{2e^2+1}{3}

Rishabh Jain - 3 years, 6 months ago

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Samarth Agarwal - 3 years, 6 months ago

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

Samarth Agarwal - 3 years, 6 months ago

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Nope... :-).... I'm not sure but it might be an IIT JEE problem.... I found this problem interesting so I found it worth sharing.

Rishabh Jain - 3 years, 6 months ago

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

Samarth Agarwal - 3 years, 6 months ago

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Sometimes I don't get differential equations, what should I do?

Akshay Yadav - 3 years, 6 months ago

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@Akshay Yadav Only practice helps in this topic..... And there are some standard rules which are needed to be followed

Samarth Agarwal - 3 years, 6 months ago

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Can i post the next question?

Adarsh Kumar - 3 years, 6 months ago

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@Adarsh Kumar Yeah! Anybody is welcome to do so.

Akshay Yadav - 3 years, 6 months ago

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@Akshay Yadav Can we change the note after 40 problems as my phone has already started lagging

Samarth Agarwal - 3 years, 6 months ago

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@Samarth Agarwal Same is happening with my phone too...

Rishabh Jain - 3 years, 6 months ago

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@Samarth Agarwal My tablet too is going crazy. I will change it tomorrow (even if the question number doesn't crosses 40). Till then let's hope it the question number crosses 40!

Akshay Yadav - 3 years, 6 months ago

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@Adarsh Kumar Yes if @Akshay Yadav allows

Samarth Agarwal - 3 years, 6 months ago

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Problem 38\text{Problem 38}What is the value of the integral given below?tanxa+btan2xdx\int\dfrac{\tan x}{a+b\tan^2x}dx

Adarsh Kumar - 3 years, 6 months ago

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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

Samarth Agarwal - 3 years, 6 months ago

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I=tanxa+b(tan)2x =sin(x)cos(x)acos2x+bsin2x\large I = \int \frac{tanx}{a+b(tan)^2x}\ = \int \frac{sin(x)cos(x)}{acos^2x+bsin^2x}

Put sin2x=t\text{Put } sin^2x=t

I=12sin(2x)asin2x(ab)=dt2(at(ab))=12(ba)lnat(ab)+CI = \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

Aditya Narayan Sharma - 3 years, 6 months ago

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Problem no : 32

f(x)=3ex+11+x2 f(x) = 3e^{x} + \dfrac{1}{1+x^{2}}
Let g(x) g(x) be the inverse of f(x) f(x) .
Compute,
33e12g(x+1)dx \displaystyle \int_{3}^{3e - \frac{1}{2}}g(x+1)dx

A Former Brilliant Member - 3 years, 6 months ago

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I=33e12g(x+1)dx=43e+12g(x)dx=f(0)f(1)g(x)dxI = \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)=tf(g(x))=f(t) g(x) = t \rightarrow f(g(x)) = f(t)
But since, f(x) f(x) and g(x) g(x) are inverses of each other, f(g(x))=x f(g(x))= x
x=f(t)dx=f(t)dt \therefore x = f(t) \rightarrow dx = f'(t)dt
I=01tf(t)dt \therefore I = \displaystyle \int_{0}^{1} tf'(t)dt
Integrating by parts,
I=[tf(t)]0101f(t)dt I = \left[tf(t)\right]_{0}^{1} - \displaystyle \int_{0}^{1} f(t)dt
I=3e+12013et+11+t2dt=3e+12[3et+tan1(x)]01 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}
I=3e+123eπ4+3+0=72π4=14π4 \therefore I = 3e + \dfrac{1}{2} -3e - \dfrac{\pi}{4} + 3 + 0 = \dfrac{7}{2} - \dfrac{\pi}{4} = \dfrac{14-\pi}{4}

A Former Brilliant Member - 3 years, 6 months ago

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PROBLEM 34:

Find the minimum value of the integral 0π(xπabπcosx)2 dx\displaystyle \int_0^{\pi} \left(x - \pi a - \frac {b}{\pi}\cos x\right)^2 \ dx

Nihar Mahajan - 3 years, 6 months ago

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

0π(x2+π2a2+b2π2cos2(x)2πax+2abcos(x)2bπcos(x))dx\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-

3b2+24b+6π4a26π4a+2π46π\frac{3b^2+24b+6{\pi}^4a^2-6{\pi}^4a+2{\pi}^4}{6{\pi}}

Let this equals to yy and then we use partial differentiation,

From that we get a=12a=\frac{1}{2} and b=4b=-4.

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

π3128π\boxed{\frac{\pi^3}{12}-\frac{8}{\pi}}

Akshay Yadav - 3 years, 6 months ago

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PROBLEM 35:

Time for an easy problem!

Find the radius and coordinates of center of curvature of the curve

y=(4xx23)12y=(4x-x^2-3)^{\frac{1}{2}}

at the points x=1.2x=1.2 and x=2x=2. Also what is this curve?

Akshay Yadav - 3 years, 6 months ago

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

Samarth Agarwal - 3 years, 6 months ago

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I think that question is more based on co ordinate geometry as curve is just a semi circle

Samarth Agarwal - 3 years, 6 months ago

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Wow, I had expected a fast reply but you did it extremely fast!

Akshay Yadav - 3 years, 6 months ago

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@Akshay Yadav This is because it just requires simple co ordinate geometry

Samarth Agarwal - 3 years, 6 months ago

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PROBLEM 36

Let f(x)f(x) be a differentiable function and f(x)=f(x),f(x)=g(x)f'(x)=-f(x) , f'(x) =g(x)

Let h(x)=[f(x)]2+[g(x)]2h(x)=[f(x)]^2+[g(x)]^2 and h(5)=11 find h(10)

Samarth Agarwal - 3 years, 6 months ago

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

Integrate both sides to get: f(x)=Cexf(x)=Ce^{-x}     g(x)=f(x)=Cex\implies g(x)=f'(x)=-Ce^{-x}     h(x)=2C2e2x\implies h(x)=2C^2e^{-2x}

h(10)h(5)=e20e10=e10\dfrac{h(10)}{h(5)}=\dfrac{e^{-20}}{e^{-10}}=e^{-10}

    h(10)=11e10\implies h(10)=\boxed{11e^{-10}}

Is it or not???

Rishabh Jain - 3 years, 6 months ago

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Absolutely correct!!! You may post the next question :)

Samarth Agarwal - 3 years, 6 months ago

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Problem39\mathbf{Problem} \mathfrak{39} Determine : dxsecx+2sinx\int \frac{dx}{secx+2sinx}

Aditya Narayan Sharma - 3 years, 6 months ago

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The given integral can be written as

cosx dx1+2sinxcosx\displaystyle \int \dfrac{\cos x \ dx}{1+2 \sin x \cos x}

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

=12(sinx+cosx)+(cosxsinx)(sinx+cosx)2dx\displaystyle = \dfrac{1}{2} \int \dfrac{(\sin x + \cos x) + (\cos x - \sin x)}{(\sin x + \cos x )^2} dx

=12dxsinx+cosx+12d(sinx+cosx)(sinx+cosx)2 \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}

=122csc(x+π4)dx+12d(sinx+cosx)(sinx+cosx)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}

=122lncsc(x+π4)+cot(x+π4)12(sinx+cosx)+C \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

Samuel Jones - 3 years, 6 months ago

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Yeah same form as @Adarsh Kumar , He just simplified the last one to 1221sin(x+π4)\frac{1}{2\sqrt2}\frac{1}{sin(x+\frac{\pi}{4})}

Aditya Narayan Sharma - 3 years, 6 months ago

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@Aditya Narayan Sharma I think @Akshay Yadav now the note should be changed and