- #1
WMDhamnekar
MHB
- 376
- 28
It is possible to find positive integers $A,B, C, D, E$ such that
$\displaystyle\int_0^{\frac{2a}{a^2+1}} sin^{-1}\big(\frac{|1-ax|}{\sqrt{1-x^2}}\big)dx=\frac{A}{\sqrt{a^2+1}}sin^{-1}\big(\frac{1}{a^B}\big ) - C sin^{-1} \big(\frac{1}{a^D}\big) + \frac {Ea\pi}{a^2+1}$ for all real numbers $ a \geq 3$.
What is the value of A + B + C + D + E ?
Answer:-
How to answer this question?. The questioner also provided answer to this question. First he solved this question using integration by parts, then applying the substitution $(a^2+1)*x- a= a*sin\theta $. Then applying the substitution $t= tan \frac{\theta}{2} $
But I didn't understand some steps in his answer.
If any member of MHB knows the solution to this question, he may reply with correct answer.
$\displaystyle\int_0^{\frac{2a}{a^2+1}} sin^{-1}\big(\frac{|1-ax|}{\sqrt{1-x^2}}\big)dx=\frac{A}{\sqrt{a^2+1}}sin^{-1}\big(\frac{1}{a^B}\big ) - C sin^{-1} \big(\frac{1}{a^D}\big) + \frac {Ea\pi}{a^2+1}$ for all real numbers $ a \geq 3$.
What is the value of A + B + C + D + E ?
Answer:-
How to answer this question?. The questioner also provided answer to this question. First he solved this question using integration by parts, then applying the substitution $(a^2+1)*x- a= a*sin\theta $. Then applying the substitution $t= tan \frac{\theta}{2} $
But I didn't understand some steps in his answer.
If any member of MHB knows the solution to this question, he may reply with correct answer.