Thread to share Good Questions from Calculus which requires deep thinking


5000 for this course (per month) for calculus.
But₹30k for 1year(discount)
Isn't it better than the big coachings

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I paid 2.1 l for 4 years. Not a great deal if you ask me and that too only for calculus. Even MC sir etoos costs less per year i suppose.

Also it’s a one time swipe bro... 30k on credit card means almost that months limit won’t be far off. Always large transactions should take place through cheques or EMI.



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Take the whole product as y
Take log on both the sides ...
Now expand the log on the product sides by using expansion of log(1- x)
Now use sum of the series as integral


After expanding got it as
{ 1 \over n }× \sum \frac {1}{1+ar} -{1 \over 2} \frac {1}{{{(1+ar)}}^2}......
Not able to get r \over n factor

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Please do not make off topic posts.
Start a new thread instead.

Final warning.


Acknowledged. @goiit-admin

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Make it r \over n
Multiply and divide the ar Wala term by n
So we get the denominator as (1+anx)
So the integral is 1 \over anx dx
The squared and the cubed terms will all be zero ...
And the first integral which will be in log can also be proved to be zero
Here summation of infinite terms will yield zero as the terms are exponentially decreasing so we can safely assume them zero

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But afaik
The n wala term should vanish in the integral and integration should come independent of n in series summation...
I haven't did any ques in which n resides behind

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Did it without gamma or digamma with simple reduction..
\int _0 ^1 1×(1-x^a)^n
Taking 1 as second function and IBP
\int _0 ^1 x(1-x^a)^n - \int _0 ^1 xn(1-x^a)^{n-1} ×ax^{a-1} × {-1}
First term cancels ..
-na × \int _0 ^1 (1-x^a)^{n-1} ×(1- x^a -1)
=-naf_n +anf_{n-1}
f_n= \frac {an}{1+an}f_{n-1}
Actually i only know gamma function for only integral of sin^nx×cos^m x

Doubt on integrations

Nice @arush_kumar_singh bro

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If I remember correctly it's the "Ahmed's integral" right?


For first one
{πln2}- { π \over 2 } { \sum _1 ^{ \infty } \frac{{2r-1})}{r.r! × 2^r }}
:sweat: Sum is not striking bro..
I think I have seen it somewhere


Is the answer 0? (for first one) I.e post no. 295


Let : \displaystyle \color{red}\sqrt{4x-x^2} = xt\Rightarrow x=\frac{4}{1+t^2} .

Then \displaystyle I=\int_0^\infty \ln\left(\frac{4}{1 +t^2} \right)\cdot \frac{1 +t^2} {4 t}\cdot \frac{8t}{(1+t^2) ^2} dt

So we have \displaystyle =2\int_0^\infty\frac{\ln 4-\ln(1+t^2)} {1 +t^2} dt

Split in two integrals, and for the second one let t=\tan y

then use well known integral \displaystyle \int_0^\frac{\pi}{2} \ln(\cos x) dx=-\frac{\pi} {2} \ln 2

So finally we get \displaystyle \color{blue}I=2\pi \ln2-2 \pi \ln2=0.


@arush_kumar_singh I was also stuck up there and so to cross check I took the help of actual limit i.e. y
For a=2 , y can be written as \over
Took the approximate value of n! (Sterling approximations) to say that y^ \frac{1}{n} is 1

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@Rohan_Shinde1 bro can you please send the solution to this one rather a hint to solve I am stuck up at tanx=x



For the first one here is my method :

I=\int_0^4 \frac {\ln x}{\sqrt {x(4-x)}} dx =\frac 12\int_0^4 \frac {\ln ((4-x)x)}{\sqrt {x(4-x)}} dx


J(a)=\int_0^4 x^a(4-x)^a dx

What we need is \frac 12 J'\left(-\frac 12\right)

For J(a) substitute x=4t to change the integral to
J(a)=4^{2a+1}\int_0^1 t^a(1-t)^a dt=4^{2a+1} B(a+1,a+1)

Hence J'(a) =4^{2a+1}B(a+1,a+1)\left(2\ln 4+2\psi(a+1)-2\psi(2a+2)\right)
J'\left(-\frac 12\right)=B\left(\frac 12,\frac 12\right)\left(2\ln 4-2\psi \left(\frac 12\right)-2\psi(1)\right)

Now just use that \psi\left(\frac 12\right)=-\gamma -\ln 4 and \psi(1)=-\gamma to get the answer.

Here B(x,y) is standard Beta function, \gamma is the Euler-Mascheroni constant and \psi(z) is the Digamma function.