Some people love solving math just for its own sake – they find beauty in mathematics. For them, math is not only used for application in physics or engineering. It’s just beautiful in itself – on its own – PURE, BEAUTIFUL.

The “beauty” can be found in equations, expressions and the flow that takes place when you form a solution. Such beauty can be found in the following questions.

If you’re a Calculus student or a math enthusiast – try out these 5 beautiful questions from Integral Calculus. I made these questions a long time ago and have received solutions for only two of them so far. See if you can solve them all.

Question #1 Question #2 Question #3 Question #4 Question #5 You can leave your answers as a comment here or mail them to me at devrishabh@gmail.com

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1. Tell the ans of ques #4 I have doubt!!!

2. Is the 2nd question arccos or (cosx)^-1?

3. For question #4 hv tried by simplifying the given integral up to integral{[Cos(taninv(1/square root1 x^c)]^2dx.again simplifying the expression tointegral{(1 x^2)/(2 x^2)}dx by letting tanm=1/sq rt(1 x^2) then (cos m)^2 =(1 x^2)/(2 x^2).hence(cos(taninv1/sqr1 x^2)=(1 x^2)/(2 x^2).finally integrating the simplified expressions i got the answer as x-0.5sqroot[taninv(x/sqroot2)] constant

4. Question can easily be solved by simplify{x(pi 49)}^15/7 to x^15/7(pi 49)^15/7 then use substituion method(by letting x^pi 7=u and differentiate to get du=pix^(pi-1)dx.but pi-1=15/7 and continue to solve it.

• Hi Salim! Thanks for reading! The questions are not meant to be difficult at all! Just plain beautiful. The flow of simplicity adds to their beauty :)

• Can you send me solution PDF please

5. Could you email the steps / worked out problems.. would like to see…

6. in question 2, the denominator should have arc sin or some other inverse trigonometric function

• Question is right

7. Hi
I can solve them.
thanks

8. @Rishabh Dev
Thanks Rishabh Dev. I have done question #1. I am quoting the first picture. What is the institution and city where the work is housed? for citing your work

9. @Rishabh Dev
Thanks Rishabh Dev. I have done question #1. I am quoting the first picture. What is the institution and city where the work is housed? for citing your work

10. Hay Rishabh Dev, can I use trigonometric idenities and formulaes from my Calculus book and other sources.

• Hello Nicolas, You can refer to any book or formulas you wish. The questions are only to enhance your ability to think creatively.

11. arka ..loves math on

ans 1 is :(7/22)* (pi+49)^15/7*(1/pi^2)* (ln|x^(1/7)|) ..trick is .. just replace pi=22/7 in case of power of x^pi in denominator

• but pi is not exactly 22/7 . pi is irrational ?????

• Nice point! Though while solving Calculus problems, we usually see pi and e from a high level view and even logarithms and other values are treated as approximations as the aim of the problem is to test calculus concepts. However, what you pointed out is right and a small instruction can be added saying “Assume pi = 22/7” however, that would also serve as a big hint which was the reason why that was avoided.

12. @John
1. {(pi + 49)^15/7 * ln (x^pi + 7)} / pi^3

13. This is really the beauty of mathematics so cute, anyway quite nice to solve.

14. ans-1.(pi+49)^pi-1*ln(x^pi+7)/pi^3

15. ans 04: x-( arc(x/sqrt 2))/(sqrt 2)

16. ans01: (〖(π+49)〗^(15/7) ln⁡(x^π+7))/π

17. ans 01: (〖(π+49)〗^2 ln⁡(x^π+7))/π

18. Trust me.. You should try them out yourself first. They may seem difficult, but they’re really not..

19. They’re really awesome.. Where can i find the solutions??