Thread: Using advanced calculus and trigonometry, for finding values

1. Using advanced calculus and trigonometry, for finding values

It is possible to find positive integers A,B,C,D,E such that $\displaystyle\int_0^{\frac{2a}{a^2+1}}\sin^{-1}\left( \frac{|1-a*x|}{\sqrt{1-x^2}} \right) dx= \frac{A}{\sqrt{a^2+1}} \sin^{-1}\left( \frac{1}{a^B} \right) - C \sin^{-1}\left( \frac{1}{a^D} \right) + \frac{E*a*\pi}{a^2+1}$

for all real numbers $a \geq 3$. What is the value of A + B + C + D+ E ?

Answer :- Answer to this question has been provided on another math and science website on internet, but I didn't understand some of the steps in that answer. I shall reproduce that answer here under this thread for clarifications on those unclear steps included in the answer within a short period of time.

Meanwhile if someone knows the answer to this question, he may reply with correct answer.

2. Re: Using advanced calculus and trigonometry, for finding values Originally Posted by WMDhamnekar It is possible to find positive integers A,B,C,D,E such that $\displaystyle\int_0^{\frac{2a}{a^2+1}}\sin^{-1}\left( \frac{|1-a*x|}{\sqrt{1-x^2}} \right) dx= \frac{A}{\sqrt{a^2+1}} \sin^{-1}\left( \frac{1}{a^B} \right) - C \sin^{-1}\left( \frac{1}{a^D} \right) + \frac{E*a*\pi}{a^2+1}$

for all real numbers $a \geq 3$. What is the value of A + B + C + D+ E ?

Answer :- Answer to this question has been provided on another math and science website on internet, but I didn't understand some of the steps in that answer. I shall reproduce that answer here under this thread for clarifications on those unclear steps included in the answer within a short period of time.

Meanwhile if someone knows the answer to this question, he may reply with correct answer.
Hello,

First by integrating by parts, then applying the substitution $(a^2+1)*x-a= a*sin \theta$, and then applying the substitution $t= tan \frac{\theta}{2}$ gives

$\large F(x)= \displaystyle\int \sin^{-1}{\left(\frac{1-a*x}{\sqrt{1-x^2}}\right)}dx$

$$=\large x \sin^{-1}{\left(\frac{1-a*x}{\sqrt{1-x^2}}\right)} + \displaystyle\int \frac {x*(a-x)}{(1-x^2)*\sqrt{2*a*x-(a^2+1)*x^2}}dx$$

$$=\large x \sin^{-1}{\left(\frac{1-a*x}{\sqrt{1-x^2}}\right)} + \sqrt{a^2+1} \displaystyle\int \frac{x*(a-x)}{(1-x^2) \sqrt{a^2-((a^2+1)*x-a)^2}} dx$$

$$=\large x \sin^{-1}{\left(\frac{1-a*x}{\sqrt{1-x^2}}\right)}+ \frac{a^2}{a^2+1} \displaystyle\int \frac {(1+ \sin {\theta}*(a^2-\sin{\theta})}{(a^2-a+1-a*\sin{\theta})*(a^2+a+1+a*\sin{\theta})} d\theta$$

$$=\large x \sin^{-1}{\left(\frac{1-a*x}{\sqrt{1-x^2}}\right)} + \frac {a^2}{a^2+1} \displaystyle\int \left(\frac{1}{a^2} + \frac{(a^2+1)*(a-1)}{2a^2*(a^2-a+1-a*\sin{\theta})} - \frac{(a^2+1)(a+1)}{2a^2*(a^2+a+1+a*\sin{\theta})} \right) dt$$ how $d\theta$ is replaced by dt? $dt=\frac{1+t^2}{2}d\theta$ This step involves partial fraction of previous integrand.

$$= \large x \sin^{-1}{\left(\frac{1-a*x}{\sqrt{1-x^2}}\right)} + \frac {\theta}{\sqrt{a^2+1}} + \sqrt{a^2+1} \displaystyle\int \left( \frac{a-1}{(a^2-a+1)*(t^2+1)- 2at} - \frac{a+1}{(a^2+a+1)(t^2+1)+2at)}\right) dt$$ This step is difficult to understand. Hence clarification is required.

$$=\large x \sin^{-1}{\left(\frac{1-a*x}{\sqrt{1-x^2}}\right)} + \frac {\theta}{\sqrt{a^2+1}} + \tan^{-1}\left(\frac{(a^2-a+1)*t-a}{(a-1)\sqrt{a^2+1}}\right) - \tan^{-1}\left(\frac{(a^2+a+1)*t + a}{(a+1)\sqrt{a^2+1}}\right)$$

For $0 \leq \large x \leq \frac {2a}{a^2+1}$ , ignoring the constant of integration, thus

$$\displaystyle\int_0^{\frac{2a}{a^2+1}} \sin^{-1} {\left(\frac{|1-a*x|}{\sqrt{1-x^2}}\right)} dx = 2F(a^{-1})- F(0) -F\left(\frac{2a}{a^2+1}\right)$$ How to interpret this step?

If $\large x=0$ we have $\theta =-\frac{\pi}{2}$ and t=-1, so that

$$F(0)=-\frac{\pi}{2*\sqrt{a^2+1}} + \tan^{-1} \left(-\frac{\sqrt{a^2+1}}{a-1}\right) - \tan^{-1}\left(-\frac{\sqrt{a^2+1}}{a+1}\right) = -\frac{\pi}{2\sqrt{a^2+1}}- \tan^{-1}\left(\frac{\sqrt{a^+1}}{a^2}\right)$$

If $\large x= \left(\frac{2a}{a^2+1}\right) , \theta =\frac{\pi}{2}, t=1$, so that,

$$F\left(\frac{2a}{a^2+1}\right)=-\frac{a\pi}{a^2+1} + \frac{\pi}{2\sqrt{a^2+1}} + \tan^{-1}\left(\frac{a-1}{\sqrt{a^2+1}}\right) - \tan^{-1}\left(\frac{a+1}{\sqrt{a^2+1}}\right)$$

$$= -\frac{a\pi}{a^2+1} + \frac{\pi}{2\sqrt{a^2+1}} -\tan^{-1} \left(\frac{\sqrt{a^2+1}}{a^2}\right)$$

Finally if $\large x= a^{-1} , \theta =\sin^{-1}{(a^{-2})}$ and hence $t=a^2-\sqrt{a^4-1}$. If $a\geq 3$ then

$$(a^4+a^2+1)\sqrt{a^2-1}-a^4\sqrt{a^2-1} > 0$$ How to interpret this step?

(in the following computation, It is difficult to understand the use of trigonometric identities. So clarifications are needed.)

And we can show that

$$F(a^{-1})=\frac{1}{\sqrt{a^2+1}} \sin^{-1}{(a^{-2})} + \tan^{-1}\left(\frac{a\sqrt{a^4-1}- (a^3+1)}{\sqrt{a^2-1}}\right) - \tan^{-1}\left(\frac{a\sqrt{a^4-1}-(a^3-1)}{\sqrt{a^2-1}}\right)$$

$$=\frac{1}{\sqrt{a^2+1}}\sin^{-1}{(a^{-2})} - \tan^{-1}\left(\frac{1}{(a^4+a^2+1)\sqrt{a^2-1}-a^4\sqrt{a^2-1}}\right)$$

$$= \frac{1}{\sqrt{a^2+1}} \sin^{-1}{(a^{-2})} -\frac{\pi}{2} + \tan^{-1}\left((a^4+a^2+1)\sqrt{a^2-1}-a^4\sqrt{a^2-1})\right)$$

and hence

$$\displaystyle\int_0^{\frac{2a}{a^2+1}} \sin^{-1}\left(\frac{|1-a*x|}{\sqrt{1-x^2}}\right)dx = \frac{2}{\sqrt{a^2+1}}\sin^{-1}{(a^{-2})}- \pi + \frac{a\pi}{a^2+1} + 2\tan^{-1}\left((a^4+a^2+1)\sqrt{a^2-1}-a^4\sqrt{a^2-1})\right) + 2 \tan^{-1} \left(\frac{\sqrt{a^2+1}}{a^2}\right)$$

$$=\frac{2}{\sqrt{a^2+1}} \sin^{-1}{(a^{-2})} -\pi +\frac{a\pi}{a^2+1} + 2\tan^ {-1}\left(\frac{1}{\sqrt{a^2-1}}\right)$$

$$=\frac{2}{\sqrt{a^2+1}}\sin^{-1}{(a^{-2})} -\pi +\frac{a\pi}{a^2+1} + 2 \cos^{-1}{(a^{-1})}$$

$$=\frac{2}{\sqrt{a^2+1}}\sin^{-1}{(a^{-2})}- 2 \sin^{-1}{(a^{-1})} + \frac{a\pi}{a^2+1}$$

So that,

A=2, B=2,C=2, D=1, E=1 and hence A+B+C+D+E=8

3. Re: Using advanced calculus and trigonometry, for finding values Originally Posted by WMDhamnekar It is possible to find positive integers A,B,C,D,E such that $\displaystyle\int_0^{\frac{2a}{a^2+1}}\sin^{-1}\left( \frac{|1-a*x|}{\sqrt{1-x^2}} \right) dx= \frac{A}{\sqrt{a^2+1}} \sin^{-1}\left( \frac{1}{a^B} \right) - C \sin^{-1}\left( \frac{1}{a^D} \right) + \frac{E*a*\pi}{a^2+1}$

for all real numbers $a \geq 3$. What is the value of A + B + C + D+ E ?

Answer :- Answer to this question has been provided on another math and science website on internet, but I didn't understand some of the steps in that answer. I shall reproduce that answer here under this thread for clarifications on those unclear steps included in the answer within a short period of time.

Meanwhile if someone knows the answer to this question, he may reply with correct answer.
Just sayin', some of the helpers on here are female!!

4. Re: Using advanced calculus and trigonometry, for finding values Originally Posted by Debsta Just sayin', some of the helpers on here are female!!
Yeah. But you're an Australian... (I bet I'm going to pay for that one someday!)

-Dan

5. Re: Using advanced calculus and trigonometry, for finding values Originally Posted by Debsta Just sayin', some of the helpers on here are female!!
Any member knowing the correct answer to my questions may reply.

Now coming to the point, how does $-\pi + \tan^{-1}{\left(\frac{1}{\sqrt{a^2-1}}\right)}= -\pi + \cos^{-1}{(a^{-1})}?$ How did the answerer compute step number 6(involving $\theta$ in numerator)?

6. Re: Using advanced calculus and trigonometry, for finding values Originally Posted by WMDhamnekar Any member knowing the correct answer to my questions may reply.

Now coming to the point, how does $-\pi + \tan^{-1}{\left(\frac{1}{\sqrt{a^2-1}}\right)}= -\pi + \cos^{-1}{(a^{-1})}?$ How did the answerer compute step number 6(involving $\theta$ in numerator)?
There are two small typos in the answer given in #2 under this thread. While computing $F(a^{-1})$ one should read the rightmost term of the R.H.S of the formula for $F(a^{-1})$ as $\tan^{-1} \left((a^4+a^2+1)\sqrt{a^2-1}- a^4\sqrt{a^2+1}\right)$

Second typo is $2 \tan^{-1}\left((a^4+a^2+1)\sqrt{a^2-1}- a^4\sqrt{a^2+1}\right) +2\tan^{-1}\left(\frac{\sqrt{a^2+1}}{a^2}\right)=2\tan^{-1}(\sqrt{a^2-1})=2\cos^{-1}(a^{-1})$

Now there are no clarifications needed as all my queries are solved by the answerer.