Search found 102 matches

by jacks
Tue Jul 12, 2016 4:12 am
Forum: Calculus
Topic: 2 Indefinite Integrals
Replies: 4
Views: 3281

2 Indefinite Integrals

(1) Evaluation of \(\displaystyle \int\frac{3x^2+1}{(x^2-1)^3}dx\)

(2) Evaluation of \(\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx\) and \(\displaystyle \int\frac{\cos^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx\)
by jacks
Mon Jul 11, 2016 6:39 pm
Forum: Real Analysis
Topic: Limit
Replies: 0
Views: 1447

Limit

If \(\displaystyle a\in \mathbb{R}\) and \(\displaystyle a\neq -1\). Then \(\displaystyle \lim_{n\rightarrow \infty}\frac{\left(1^a+2^a+\cdots+n^a\right)}{(n+1)^{a-1}\cdot \left[(na+1)+(na+2)+\cdots+(na+n)\right]} = \frac{1}{60}\) Then value of \(a\) is (which one is Right) ? \(\displaystyle (a)\;\;...
by jacks
Mon Jul 11, 2016 6:35 pm
Forum: Real Analysis
Topic: Putnam 1987 : B1
Replies: 1
Views: 1749

Re: Putnam 1987 : B1

Given \(\displaystyle \int_{2}^{4}\frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\) Let \(\displaystyle I = \int_{2}^{4}\frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} = \int_{2}^{4}\frac{\sqrt{\ln(9-(2+4-x))} dx}{\sqrt{\ln(9-(2+4-x))}+\sqrt{\ln((2+4-x)+3)}}\) Using \(\displays...
by jacks
Mon Jul 11, 2016 6:31 pm
Forum: Real Analysis
Topic: \( \int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{\left ( x^3+1 \right )\left ( x^2+1 \right )} \)
Replies: 1
Views: 1877

Re: \( \int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{\left ( x^3+1 \right )\left ( x^2+1 \right )} \)

Let \(\displaystyle I = \int_{\alpha^{-1}}^{\alpha}\frac{1}{(x^3+1)(x^2+1)}dx....................................................(1)\), Where \(\displaystyle \alpha = \sqrt{3}\) Now Let \(\displaystyle x=\frac{1}{t}\) and \(\displaystyle dx = -\frac{1}{t^2}\) and changing Limits \(\displaystyle I = ...
by jacks
Mon Jul 11, 2016 6:17 pm
Forum: Real Analysis
Topic: \( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)
Replies: 5
Views: 3774

Re: \( \int_{0}^{1}\frac{1+x^a}{\left ( 1+x \right )^{a+2}}\,dx \)

Given \(\displaystyle \int_{0}^{1}\frac{1+x^a}{(1+x)^{a+2}}dx = \int_{0}^{1}\left(1+x\right)^{-a-2}dx+\int_{0}^{1}\frac{x^a}{(1+x)^{a+2}}dx\) Now Let \(\displaystyle I = \int_{0}^{1}(1+x)^{-a-2}dx = -\frac{1}{(a+1)}\cdot \frac{1}{2^{a+1}}+\frac{1}{a+1}\) Similarly Let \(\displaystyle J = \int_{0}^{1...
by jacks
Mon Jul 11, 2016 9:23 am
Forum: Calculus
Topic: 2 Trigonometric Integral
Replies: 0
Views: 1468

2 Trigonometric Integral

\(a\): Evaluation of $$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\pi+4x^3}{2-\cos\left(|x|+\frac{\pi}{3}\right)}dx$$

\(b\):: Evaluation of \[\displaystyle \int_{0}^{\pi}e^{|\cos x|}\left(2\sin \left(\frac{\cos x}{2}\right)+3\cos \left(\frac{\cos x}{2}\right)\right)dx\]
by jacks
Mon Jul 11, 2016 9:20 am
Forum: Real Analysis
Topic: \( \int_{-\infty}^\infty \frac{dx}{\left(4+x^2 \right) \sqrt{4+x^2}} \)
Replies: 1
Views: 1846

Re: \( \int_{-\infty}^\infty \frac{dx}{\left(4+x^2 \right) \sqrt{4+x^2}} \)

Given \(\displaystyle \int_{-\infty}^{\infty}\frac{1}{(x^2+4)\sqrt{x^2+4}}dx = 2\int_{0}^{\infty}\frac{1}{(x^2+4)\sqrt{x^2+4}}dx\) Now Let \(\displaystyle x= \frac{1}{t}\) and \(\displaystyle dx = -\frac{1}{t^2}dt\) and changing Limits , So integral convert into \(\displaystyle 2\int_{0}^{\infty}\fr...
by jacks
Sun Jul 10, 2016 6:58 pm
Forum: Calculus
Topic: Definite Integrals
Replies: 2
Views: 2282

Definite Integrals

Evaluation of definite Integrals

\((a)::\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n}(x)\cdot \sin (nx)dx\)

\((b)::\displaystyle \int_{0}^{\frac{\pi}{2}}\cos^{n}(x)\cdot \cos (nx)dx\,,\) where \(n\in \mathbb{N}\)
by jacks
Sun Jul 10, 2016 4:52 pm
Forum: Calculus
Topic: 2 Integrals
Replies: 2
Views: 2250

2 Integrals

Evaluation of \(\displaystyle \int\frac{x^{5n-1}+2x^{4n-1}}{\left(x^{2n}+x^n+1\right)^3}dx\) and \(\displaystyle \int_{0}^{\pi}\frac{\sin^2(nx)}{\sin^2(x)}dx\,,\) where \(n\in \mathbb{N}\,.\)
by jacks
Sun Jul 10, 2016 4:25 pm
Forum: Calculus
Topic: Four limits
Replies: 4
Views: 3232

Re: Four limits

For \((d)\) Given \(\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\ldots+\frac{n}{n^2}\right)\) Now Convert the Infinite Series into Rein man Sum \(\displaystyle =\lim_{n\rightarrow \infty}\frac{1}{n}\cdot \left(\frac{1}{n}+\frac{2}{n}+\ldots+\frac{n}{n}\right) =\lim_{n\r...