Actually the real source of that question is MIT integration bee.

# Thread to share Good Questions from Calculus which requires deep thinking

**Jagdish_Singh**#25

Given \displaystyle I = \int^{\infty}_{0}f(x)dx, Put \displaystyle x = \frac{1}{t} and \displaystyle dx = -\frac{1}{t^2}dt

and changing limits, We get

\displaystyle I = \int^{\infty}_{0}f\bigg(\frac{1}{t}\bigg)\cdot \frac{1}{t^2}dt = \int^{\infty}_{0}f\bigg(\frac{1}{x}\bigg)\cdot \frac{1}{x^2}dx.

**Jagdish_Singh**#26

Let \displaystyle I = \int \sin (101x)\cdot \sin^{99}(x)dx = \int \sin (100x+x)\cdot \sin ^{99}(x)dx

\displaystyle I = \int \left[\sin (100x)\cdot \cos x+\cos (100x)\cdot \sin x\right]\cdot \sin^{99}(x)dx

\displaystyle I = \int \sin (100x)\cdot \sin^{99}(x)\cdot \cos (x)dx+\int \cos (100x)\cdot \sin^{100}(x)dx

Now Using \bf{I.B.P\;,} We get

\displaystyle I = \sin (100x)\cdot \frac{\sin ^{100}(x)}{100}-\int \cos(100x)\cdot 100\cdot \frac{\sin ^{100}(x)}{100}dx+\int \cos (100x)\cdot \sin^{100}(x)dx

\displaystyle I =\int \sin (101x)\cdot \sin^{99}(x)dx = \sin (100x)\cdot \frac{\sin ^{100}(x)}{100}+\mathcal{C}

**Jagdish_Singh**#27

**Using Complex number**

\displaystyle \int \sin (101x)\sin^{99}xdx = \Im\int e^{i(101x)}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{99}dx

Where \Im means Imaginary part of Integral

Now Let \displaystyle I = \int e^{i(101x)}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{99}dx = \frac{i}{2^{99}}\int e^{2ix}(e^{2ix}-1)^{99}dx

Now Put e^{2ix}-1 = u\;, and 2ie^{2ix}dx = dt

Then \displaystyle I = \frac{1}{2^{100}}\int (u)^{99}du = \frac{1}{100\cdot 2^{100}}(e^{2ix}-1)^{100}+\mathcal{C}

Now factoring out e^{100ix} and divide and multiply by (2i)^{100}\;, We get

\displaystyle I = \frac{e^{100ix}\sin^{100}(x)}{100}+\mathcal{C}\Rightarrow \Im(I)=\frac{\sin (100x)\sin^{100}x}{100}

So \displaystyle \int \sin (101x)\sin^{99}xdx = \frac{\sin (100x)\sin^{100}x}{100}+\mathcal{C}

**ananda_2018**#28

@Jagdish_Singh sir i wish to do btech.But at the same time i would like to extend my knowledge in maths and physics just for the sake of gaining knowledge till the day i exist on this planet(not wishing to take them as career though).Yur solutions are just world class.

Any tips from yur side for me to gain more high level knowledge in maths(as yu r a maths expert)?

**Jagdish_Singh**#29

@Anand_Krishnan Read more and more books of Mathematics

(Basics to Higher level). I am not a maths experts

Me also learning Higher Level Mathematics

But I dont have time due to my hectic Schedule of Classes.

I have seen that you are very Good in Mathematics. Keep it up man.

My best wishes with you.

Thanks

**ananda_2018**#30

I wont agree with yu if yu say that yu r not a maths expert...

And thnx for yur wishes @Jagdish_Singh sir.I will try my best.

Also i am learning a lot from yur solutions,those are just awesome.

**Rishi1**#34

Stuck! ln(x)/(1+x) from 0 to 1 @Jagdish_Singh @Azimuddin_Sheikh @pratyaksh_tyagi @ananda_2018

**Jagdish_Singh**#35

Let \displaystyle I = \int^{1}_{0}\frac{\ln(x)}{1+x}dx = \int^{1}_{0}\ln (x)\sum^{\infty}_{n=0}(-1)^nx^ndx

So \displaystyle I = \sum^{\infty}_{n=0}(-1)^n\int^{1}_{0}\ln(x)\cdot x^ndx

Using I .B .P Method

So \displaystyle I = \sum^{\infty}_{n=0}(-1)^n \bigg[\ln(x)\cdot \frac{x^{n+1}}{n+1}\bigg|^{1}_{0}-\frac{1}{n+1}\int \frac{1}{x}\cdot x^{n+1}dx\bigg]

So \displaystyle I = \sum^{\infty}_{n=0}(-1)^n\bigg[1-\frac{1}{n+1}\int x^{n}dx\bigg] = \sum^{\infty}_{n=0}\frac{(-1)^{n+1}}{(n+1)^2}

Now Using formula \displaystyle \bullet \; \zeta(2) = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots = \frac{\pi^2}{6}.

So \displaystyle \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots = \frac{\pi^2}{6}-\frac{1}{2^2}\cdot \frac{\pi^2}{6} = \frac{\pi^2}{8}.

So \displaystyle I = \sum^{\infty}_{n=0}\frac{(-1)^{n+1}}{(n+1)^2} = -\bigg[\frac{\pi^2}{8}-\frac{1}{4}\cdot \frac{\pi^2}{6}\bigg]=-\frac{\pi^2}{12}.

**Azimuddin_2019**#38

can anyone prove that this function must be sine? I just guess for that. But I need a proper Solution for f(x)

9th one @Jagdish_Singh @pratyaksh_tyagi @Unik_2018 @ananda_2018**Unik_2018**#39

Differentiate the eqn wrt x

F²(x) + F'²(x) = 1

F'(x) /√(1-F²(x)) = 1. [note that here will be two cases +-]

Now integrate wrt dx

sin^{-1} (f(x)) =x+C

Similarly

Case2

cos^{-1} (f(x)) =x+C

Hence proved

**ananda_2018**#41

Actually that problem in post 36 is something which we all have done while studying integration......just forget those big numbers and the problem is done.

Yep correct @Unik_2018