×

# 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.

• 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]** 

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

Note by Akshay Yadav
10 months ago

Sort by:

Problem 25

Prove that:

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

This problem has been solved by Aditya Sharma. · 10 months ago

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

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

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

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

Rearranging we get:

$\dfrac{5050I_1}{I_2}=\boxed{5051}$ · 9 months, 4 weeks ago

Both methods are same. The proof of Gamma functional equation uses IBP, so both methods are essentially the same. · 9 months, 3 weeks ago

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 ,

$$\large \frac{5050.\int_{0}^{1}(1-x^{50})^{100}dx}{\int_{0}^{1}(1-x^{50})^{101}dx} = 5051$$ $$\large \boxed{Proved}$$ · 10 months ago

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

This problem has been solved by Harsh Shrivastava. · 10 months ago

· 10 months ago

A similar problem like this came in JEE Mains 2016! I just saw the question paper and recalled this problem ;) · 9 months, 3 weeks ago

Yeah. · 9 months, 3 weeks ago

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

Prove that : $$\large \int_{0}^{\pi} \frac{xdx}{1+cos(\alpha)sin(x)} = \frac{\alpha\pi}{sin(\alpha)}$$ · 9 months, 4 weeks ago

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

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

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

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

$\large \boxed{=\dfrac{\pi\alpha}{\sin \alpha}}$ · 9 months, 4 weeks ago

Problem 30

$\large\displaystyle \int \dfrac{\mathrm{d}x}{x^{11}\sqrt{1+x^4}}=\ ?$ · 9 months, 4 weeks ago

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

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 , $$I = - \frac{ \sqrt{x^4 + 1}( 8x^8 - 4x^4 + 3)}{30x^{10}} + C$$ · 9 months, 4 weeks ago

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

$\dfrac{\displaystyle\int_0^{2016}\left(\left(\frac1{p^2(p-1)!}-\mathfrak S\right)\dfrac{p(x+p)!}{x!}\right)dx}{\displaystyle\int_0^{2}\left(\left(\frac1{p^2(p-1)!}-\mathfrak S\right)\dfrac{p(x+p)!}{(x-2)!}\right)dx}= \color{blue}{\varphi}$

Find $$\dfrac{\sqrt{60^2-\color{blue}{\varphi}}}{3!}$$.

**Original · 9 months, 4 weeks ago

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. · 9 months, 4 weeks ago

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

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

$$f(4) = \frac{5(e^8-1)}{2e^4}$$ · 9 months, 4 weeks ago

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 $$f(4) = \displaystyle \sum_{n=1}^\infty \frac{ 4^n ( (-1)^n + 5 ) }{ 8(n-1)!} = \displaystyle \sum_{n=1}^\infty \frac{ 4^n (-1)^n }{ 8(n-1)!} + \frac{5}{2} \displaystyle \sum_{n=1}^\infty \frac{ 4^{(n-1)} }{ (n-1)!} = \frac{1}{8} \frac{-4}{e^4} + \frac{5}{2} e^4 = \boxed{ \frac{ 5e^8 - 1}{2e^4} }$$ · 9 months, 4 weeks ago

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. · 9 months, 3 weeks ago

This is correct ^. In reality the function is exponential. · 9 months, 2 weeks ago

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 ,

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

$$\large\boxed{ f(4) = \frac{5(e^4-e^{-4})}{2} = \frac{5(e^8-1)}{2e^4}}$$ · 9 months, 4 weeks ago

Problem 24:

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

This problem was solved by Rishabh Cool. · 10 months ago

Solution to Problem 24

Substitute $$\ln \tan x=t$$ such that $$\dfrac{dx}{\sin 2x}=\dfrac{dt}{2}$$. Integral transforms to: $\int \dfrac{dt}{2 t^3}$ $=\dfrac{-1}{ 4t^2}=\dfrac{-1}{4\ln^2 \tan x}+C$ · 10 months ago

Problem 31

If $$I_{1} = \displaystyle \int_0^{\frac{\pi}{2} } f( \sin 2x) \sin x \mathrm{d}x$$ and $$I_{2} = \displaystyle \int_0^{\frac{\pi}{4} } f( \cos 2x) \cos x \mathrm{d}x$$ , then find $$\frac{I_{1}}{I_{2}}$$ ? · 9 months, 4 weeks ago

$$I_{1} = \displaystyle \int_{0}^{\dfrac{\pi}{2}} f(\sin 2x)\sin x dx$$
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 $$\displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b} f(a+b-x)dx$$
$$\therefore I_{2} = \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)\cos\left( \dfrac{\pi}{4} - x\right) dx$$
$$\therefore I_{2} = \dfrac{1}{\sqrt{2}}\cdot \displaystyle \int_{0}^{\dfrac{\pi}{4}} f(\sin 2x)(\cos x + \sin x) dx = \dfrac{I_{1}}{\sqrt{2}}$$
$$\dfrac{I_{1}}{I_{2}} = \sqrt{2}$$ · 9 months, 4 weeks ago

Problem 29:

Prove that:

$\displaystyle \lim_{x\to 0} \dfrac{24}{x^3} \displaystyle \int_0^x \dfrac{t \ln(1+t)}{t^4+4} dt =2$ · 9 months, 4 weeks ago

Applying LHopital:

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

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

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

$=\boxed 2$ · 9 months, 4 weeks ago

$$\text{Problem 38}$$What is the value of the integral given below?$\int\dfrac{\tan x}{a+b\tan^2x}dx$ · 9 months, 2 weeks ago

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 · 9 months, 2 weeks ago

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

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

$$I = \frac{1}{2}\int \frac{sin(2x)}{a - sin^2x(a-b)} = \int \frac{dt}{2(a - t(a-b))} = \frac{1}{2(b-a)}ln|a-t(a-b)| + C$$ · 9 months, 2 weeks ago

Problem 37

Let $$f(x)$$ be a differentiable function such that $$\displaystyle \dfrac{d f(x) }{dx} = f(x) + \int_0^2 f(x) \, dx$$ and $$f(0) = \dfrac{4-e^2}3$$, then prove: $f(2)=\dfrac{2e^2+1}{3}$ · 9 months, 2 weeks ago

· 9 months, 2 weeks ago

I don't have a good problem right now.... Can anybody else post the question... Thanx · 9 months, 2 weeks ago

Can i post the next question? · 9 months, 2 weeks ago

Yeah! Anybody is welcome to do so. · 9 months, 2 weeks ago

Comment deleted 9 months ago

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

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

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

Like this ? · 9 months, 2 weeks ago

Just cut and paste it to Adarsh's new comment..:-) · 9 months, 2 weeks ago

Yup ! · 9 months, 2 weeks ago

@Adarsh Kumar under the heading comments your name is written with a box below it.... U have to write in it · 9 months, 2 weeks ago

Is this fine?(i have posted) · 9 months, 2 weeks ago

Comment deleted 9 months ago

Yup!Just put that expression in mod,and add the integration constant :) · 9 months, 2 weeks ago

@Adarsh Kumar pls post it separately... Thanks · 9 months, 2 weeks ago

Separately meaning,not in response to anybody? · 9 months, 2 weeks ago

Pls post it separately... Thanks · 9 months, 2 weeks ago

Can we change the note after 40 problems as my phone has already started lagging · 9 months, 2 weeks ago

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! · 9 months, 2 weeks ago

Same is happening with my phone too... · 9 months, 2 weeks ago

Yes if @Akshay Yadav allows · 9 months, 2 weeks ago

Sometimes I don't get differential equations, what should I do? · 9 months, 2 weeks ago

Only practice helps in this topic..... And there are some standard rules which are needed to be followed · 9 months, 2 weeks ago

@Rishabh Cool is this your original problem??? Btw Its really very nice · 9 months, 2 weeks ago

Nope... :-).... I'm not sure but it might be an IIT JEE problem.... I found this problem interesting so I found it worth sharing. · 9 months, 2 weeks ago

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

The answer is very long so be patient.! · 9 months, 3 weeks ago

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

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

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

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

$$I = \dfrac{1}{2\sqrt{2}}\ln\left(\dfrac{\sqrt{\tan(x)} + \sqrt{\cot(x)}-\sqrt{2}}{\sqrt{\tan(x)} + \sqrt{\cot(x)}+\sqrt{2}}\right)+\dfrac{1}{\sqrt{2}}\cdot \arctan\left(\dfrac{\sqrt{\tan(x)}-\sqrt{\cot(x)}}{\sqrt{2}}\right) + c$$ · 9 months, 3 weeks ago

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

(Sorry to be rude) · 9 months, 3 weeks ago

You post. I have been trying to come up with a good problem, but I haven't been getting one. · 9 months, 3 weeks ago

Okay according to rules @Samarth Agarwal should post the next problem. · 9 months, 3 weeks ago

Very nice..... Pls post the next question · 9 months, 3 weeks ago

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

Find $$I_{20} - I_{6}$$ · 9 months, 3 weeks ago

Comment deleted 9 months ago

The answer is not 0.
$$I_{n} = \displaystyle \int_{0}^{\infty} \dfrac{ \displaystyle \sum_{k=1}^{n} \left(\arctan\left(\dfrac{(k+1)x}{k}\right) -\arctan(x) \right)}{x}dx$$

Consider the integral,
$$J(a) = \displaystyle \int_{0}^{\infty} \dfrac{ \arctan(ax) - \arctan(x) }{x} dx$$
$$\dfrac{dJ}{da} = \displaystyle \int_{0}^{\infty} \dfrac{\partial}{\partial a} \dfrac{\arctan(ax)-\arctan(x)}{x} dx$$
$$\dfrac{dJ}{da} = \displaystyle \int_{0}^{\infty} \dfrac{1}{1+(ax)^{2}}dx$$
$$\dfrac{dJ}{da} = \dfrac{\pi}{2a}$$
$$\therefore J = \dfrac{\pi \ln(a)}{2} + c$$
Since, $$J(1)= 0 , \rightarrow c = 0$$
$$\therefore J(a) = \dfrac{\pi \ln(a)}{2}$$
$$I_{n} = \displaystyle \sum_{k=1}^{n} J\left(\dfrac{k+1}{k}\right) = \dfrac{\pi \ln(n+1)}{2}$$
$$I_{20} - I_{6} = \dfrac{\pi}{2} \left( \ln(21) - \ln(7) \right) = \dfrac{\pi \ln(3)}{2}$$

I can differentiate under the integral when the function is continuous, and differentiable. As for why your answer is incorrect,
After substituting $$x = \dfrac{1}{t}$$
$$\arctan\left(\dfrac{(k+1)x}{k}\right) = \arctan\left(\dfrac{(k+1)}{kt}\right) = \dfrac{\pi}{2} - \arctan\left(\dfrac{kt}{(k+1)}\right) \ne \dfrac{\pi}{2} - \arctan\left(\dfrac{(k+1)x}{k}\right)$$ · 9 months, 3 weeks ago

$$\mathbf{Problem} \mathfrak{39}$$ Determine : $$\int \frac{dx}{secx+2sinx}$$ · 9 months, 2 weeks ago

The given integral can be written as

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

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

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

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

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

$$\displaystyle = -\dfrac{1}{2\sqrt{2}} \ln \left| \csc\left( x +\frac{\pi}{4}\right) + \cot \left(x+\frac{\pi}{4}\right) \right|-\dfrac{1}{2(\sin x +\cos x)} + C$$ · 9 months, 2 weeks ago

My method was a little bit different!Although nice work! · 9 months, 2 weeks ago

Yeah same form as @Adarsh Kumar , He just simplified the last one to $$\frac{1}{2\sqrt2}\frac{1}{sin(x+\frac{\pi}{4})}$$ · 9 months, 2 weeks ago

I think @Akshay Yadav now the note should be changed and the next note will begin with problem 40 as my PC,tablet,Mobile everything is at a standstill :P · 9 months, 2 weeks ago

Is the answer long? · 9 months, 2 weeks ago

Not that really. Not as that of $$\sqrt{tanx}$$ · 9 months, 2 weeks ago

Yeah not that long but does it contain log and some trigo functions? · 9 months, 2 weeks ago

Yeah it has log & trigo functions · 9 months, 2 weeks ago

Is it $\dfrac{1}{2\sqrt{2}}\left(\ln |\csc(x+\frac{\pi}{4})-\cot(x+\frac{\pi}{4})|-\dfrac{1}{\sin(x+\frac{\pi}{4})}\right)+C$ · 9 months, 2 weeks ago

Absolutely ! Great work. · 9 months, 2 weeks ago

Haha thanx!It was a good question!I am posting the solution! · 9 months, 2 weeks ago

Yeah ! · 9 months, 2 weeks ago

@Aditya Sharma @Akshay Yadav Who will post the next Problem? Me or @Adarsh Kumar He had posted the answer while I was posting the solution, but the solution was posted by me earlier. · 9 months, 2 weeks ago

I am a bit confused here, Accordingly @Samuel Jones should be allowed as he provided the solution a bit earlier. · 9 months, 2 weeks ago

Okay. Let @Akshay Yadav post the next note. I'll post Problem $$40$$ there. · 9 months, 2 weeks ago

Yeah thanx :P or else my phone would have exploded . @Akshay Yadav please post the next note :) · 9 months, 2 weeks ago

PROBLEM 36

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

Let $$h(x)=[f(x)]^2+[g(x)]^2$$ and h(5)=11 find h(10) · 9 months, 2 weeks ago

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

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??? · 9 months, 2 weeks ago

Absolutely correct!!! You may post the next question :) · 9 months, 2 weeks ago

PROBLEM 35:

Time for an easy problem!

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

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

at the points $$x=1.2$$ and $$x=2$$. Also what is this curve? · 9 months, 2 weeks ago

It is a semi circle above x axis as y is always positive with radius 1 unit and center (2,0) · 9 months, 2 weeks ago

Wow, I had expected a fast reply but you did it extremely fast! · 9 months, 2 weeks ago

This is because it just requires simple co ordinate geometry · 9 months, 2 weeks ago

I think that question is more based on co ordinate geometry as curve is just a semi circle · 9 months, 2 weeks ago

PROBLEM 34:

Find the minimum value of the integral $$\displaystyle \int_0^{\pi} \left(x - \pi a - \frac {b}{\pi}\cos x\right)^2 \ dx$$ · 9 months, 3 weeks ago

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,

$\boxed{\frac{\pi^3}{12}-\frac{8}{\pi}}$ · 9 months, 2 weeks ago

Problem no : 32

$$f(x) = 3e^{x} + \dfrac{1}{1+x^{2}}$$
Let $$g(x)$$ be the inverse of $$f(x)$$.
Compute,
$$\displaystyle \int_{3}^{3e - \frac{1}{2}}g(x+1)dx$$ · 9 months, 4 weeks ago

$$I = \displaystyle \int_{3}^{3e - \frac{1}{2}} g(x+1) dx = \int_{4}^{3e+\frac{1}{2}} g(x)dx = \int_{f(0)}^{f(1)}g(x)dx$$
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,
$$I = \left[tf(t)\right]_{0}^{1} - \displaystyle \int_{0}^{1} f(t)dt$$
$$I = 3e + \dfrac{1}{2} - \displaystyle \int_{0}^{1} 3e^{t} + \dfrac{1}{1+t^{2}}dt = 3e + \dfrac{1}{2} - \left[ 3e^{t} + \tan^{1}(x) \right]_{0}^{1}$$
$$\therefore I = 3e + \dfrac{1}{2} -3e - \dfrac{\pi}{4} + 3 + 0 = \dfrac{7}{2} - \dfrac{\pi}{4} = \dfrac{14-\pi}{4}$$ · 9 months, 3 weeks ago

Third note of this season has been posted. Click here to access it. · 9 months, 2 weeks ago