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Aug
24
comment Integral of $\arcsin$ of a rational function, using integration by parts
@kakashi, The last identity holds only if $x\le1$ See math.stackexchange.com/questions/523625/…
Aug
24
answered Alternating sum of binomial coefficients is equal to zero
Aug
24
comment How can I find $x$, $y$ values for $\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$
@Osama, Just express as $$X+iY=A+iB$$ where $X,Y,A,B$ are real
Aug
24
comment Find all ordered pairs $(a, b)$ in $a+\frac{10b}{a^2+b^2} = 5\;\;,b+\frac{10a}{a^2+b^2}=4$
We have $$10b=(5-a)(a^2+b^2), 10a=(4-b)(a^2+b^2)$$ On division, $$\frac ba=\frac{5-a}{4-b}\implies a^2-5a=b^2-4b$$ $$\iff4a^2-20a=4b^2-16b\iff(2a-5)^2-(2b-2)^2=9$$ $$\implies2a=3\sec\theta+5,2b=2+3\tan\theta$$
Aug
24
comment Why $\lim_{x\to0}\frac{\int_0^x\frac{t^2}{t^4+1}dt}{x^3}=\lim_{x\to0}\frac{\frac{x^2}{x^4+1}}{3x^2}$?
en.wikipedia.org/wiki/Differentiation_under_the_integral_sign
Aug
24
comment If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
Let us continue this discussion in chat.
Aug
24
revised If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
added 544 characters in body
Aug
24
comment Understanding proof for $\sum_{k=1}^{n-1}{\sin{\frac{2\pi k}{n}}} = 0$
@johnnycrab, See math.stackexchange.com/questions/17966/…
Aug
24
comment Polar form of the sum of complex numbers $\operatorname{cis} 75 + \operatorname{cis} 83 + \ldots+ \operatorname{cis} 147$
@John, See math.stackexchange.com/questions/17966/…
Aug
23
comment If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
@rsadhvika, Sorry for the confusion. I want to start afresh tomorrow. It's >1 AM
Aug
23
comment If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
@rsadhvika, $$\implies p(p+p'')=a^2+b^2-p^2-p'^2=?$$
Aug
23
comment If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
@rsadhvika, Yes, sorry the answer seems curt
Aug
23
answered If $ p^2=a^2 \cos^2 \theta+b^2 \sin^2 \theta $, show that $p + p'' = \frac{a^2b^2}{p^3}$
Aug
23
comment Evaluation of $ \int \left(x^{3m}+x^{2m}+x^{m}\right)\cdot \left(2x^{2m}+3x^{m}+6\right)^{\frac{1}{m}}dx\;,$
As $\dfrac{d(2x^{2m}+3x^m+6)}{dx}=mx^{-1}(4x^{2m}+3x^m)$ I was trying $$x^{3m}+x^{2m}+x^m=Ax\cdot mx^{-1}(4x^{2m}+3x^m)+Bx^{m+1}mx^{-1}(4x^{2m}+3x^m)$$
Aug
23
comment Which is greater and why?
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
Aug
23
comment Using sum/product of roots of a cubic equation to solve given expression $(a+b)^3+ (b+c)^3 + (c+a)^3$
@QuangHoang, Agreed & adjusted accordingly. Thanks for your input
Aug
23
revised Using sum/product of roots of a cubic equation to solve given expression $(a+b)^3+ (b+c)^3 + (c+a)^3$
added 52 characters in body
Aug
23
comment Using sum/product of roots of a cubic equation to solve given expression $(a+b)^3+ (b+c)^3 + (c+a)^3$
@Mathbreaker, Please find the edited version
Aug
23
comment Using sum/product of roots of a cubic equation to solve given expression $(a+b)^3+ (b+c)^3 + (c+a)^3$
@QuangHoang, Added alternative method
Aug
23
revised Using sum/product of roots of a cubic equation to solve given expression $(a+b)^3+ (b+c)^3 + (c+a)^3$
added 164 characters in body