Calculate $$\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.   $\\$     my trial  :   $~~~~~~~~~~~~~~~~~$The Elipse  $x(t) = \frac{\cos(t)}{\sqrt{3}}$ $~y(t) = \frac{\sin(t)}{\sqrt{5}}$  $t \in [0,2\pi]$  $ dx = \frac{-\sin(t)}{\sqrt{3}}~dt$  $ dy = \frac{\cos(t)}{\sqrt{5}}~dt$  $ 2xy^2~dx = \frac{2\cos(t)}{\sqrt{3}}\frac{\sin(t)^2}{5}\frac{-\sin(t)}{\sqrt{3}}~dt = \frac{-2\cos(t)\sin(t)^2\sin(t)}{15}~dt$  $ -2yx^2~dy = \frac{-2\sin(t)}{\sqrt{5}}\frac{\cos(t)^2}{3}\frac{\cos(t)}{\sqrt{5}}~dt = \frac{-2\cos(t)\cos(t)^2\sin(t)}{15}~dt$  $ x^2 + y^2 = \frac{\cos^2(t)}{3} + \frac{\sin^2(t)}{5} = \frac{3+2\cos^2(t)}{15}$  so  $~\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2} = \int_0^{2\pi}\frac{-2\sin(t)\cos(t)}{3+2\cos^2(t)}~dt = \frac{ln(3+2\cos^2(t))}{2}|_0^{2\pi} = 0$
122593
Calculate $$\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.   I tried to calculate the integral using green theorm.  now i need to build enclosier that doesn't enclose $(0,0)$ i am having hard time guessing what to build.  a circle and ellipse might be perfect but then the domain is not easy to write. can i have hint please ?
13485
I'm stuck with this integral  $\int\frac {\cos^4x}{\sin^3x} dx$    which I rewrote as  $\int \csc^3x \cos^4xdx$  then after using the half angle formula twice for $\cos^4x$ I got this  $\frac 14\int \csc^3x (1+\cos(2x))(1+\cos(2x))dx$   then after solving those products I got these integrals  $\frac 14 \{\int \csc^3xdx+2\int \csc^3x \cos(2x)dx + \int \csc^3x \cos^2(2x)dx\}$  I do know how to solve the $\int \csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!
1133
Evaluate $$\int_0^{2\pi} \frac{\sin^2\theta}{5+4\cos(\theta)}\mathrm d\theta$$   This is the final question on my review for my final exam tomorrow, and I will be honest and say that I have no clue how to begin problem. Any hints in the direction of how to solve this would be helpful.  Following from @Adam Hughes,   $-\dfrac{\pi}{2}[f'(0)+Res_2]$When I took the derivative of $f(z)=\dfrac{(z^2-1)^2}{2z^2+5z+2}$ and evaluated for 0, I got $-\dfrac{5}{4}$ Now, $Res_2$
30762
Question:     Find out $\displaystyle{\int e^{\cos x}~dx}$.   My Attempt:  Let $\cos x = y$. Hence $-\sin x\ dx = dy$ or $$dx = \displaystyle{\frac{-dy}{\sin x}=\frac{-dy}{\sqrt{1-\cos^2x}}=\frac{-dy}{\sqrt{1-y^2}}}$$ So  $$\begin{align}\int e^{\cos x}~dx &amp;= \int e^y\left(\frac{-dy}{\sqrt{1-y^2}}\right)\\ &amp;=-\int\frac{e^y}{\sqrt{1-y^2}}~dy \end{align}$$  This integral is one I can't solve. I have been trying to do it for the last two days, but can't get success. I can't do it by parts because the new integral thus formed will be even more difficult to solve. I can't find out any substitution that I can make in this integral to make it simpler. Please help me solve it. Is the problem with my first substitution $y=\cos x$ or is there any other way to solve the integral $\displaystyle{\int\frac{e^y}{\sqrt{1-y^2}}~dy}$?
5782
Calculate: $$\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)\mathrm{d}t$$ My try: $\displaystyle\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)dt\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\cos(\mathfrak{I\textrm{z}}+3(-i\log z)dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}}^{(\mathfrak{I\textrm{z}}+3(-i\log z))i}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}}z^{3}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}R}z^{3}dz$ and there is nothing here that is not holomorphic, therefore according to Cauchy theorem it must be exactly $0$.
77661
If $I_n=\int _0^{\pi }\sin^{2n}\theta \:d\theta $, show that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $  Hence calculate $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$  I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.
81431
Evaluate the line integral  $$\int_C xe^{y}\, {\rm d}s,$$  where $C$ is the line segment from $(-1,2)$ to $(1,1)$.  I do not get this part of calculus at all please show me how this is solved and if you give me hints is ok too. Please help me. Thanks
34324
How to calculate following integration?  $$\int\left( \sqrt{\tan x}+\sqrt{\cot x}\right)dx$$
94814
I am given $ f(t) = e^{\sin(t)} $ and $ F(x) = \int\limits_{0}^{x} f(t) dt $ and have to compute: $\int\limits_{-0.5  \pi}^{0.5 \pi} \sin(x) F(\cos(x)) dx  $  How do I go about this problem?   Thus far I got: $ I :=\int\limits_{-0.5  \pi}^{0.5 \pi} \sin(x) F(\cos(x)) dx = -\int\limits_{-1}^{1} F(u) du $ $ \ \ \ $(*)  And: $$ \int\limits_{-0.5  \pi}^{0.5 \pi} \sin(x) F(\cos(x)) = -cos(x) F(cos(x)) - \int\limits_{-0.5  \pi}^{0.5 \pi} \sin(x) F'(\cos(x)) \cos(x) dx = 0 + \int\limits_{-1}^{1} F'(u)u du = [F(u)u]_{-1}^{1} - \int\limits_{-1}^{1} F(u)du = 2.28319 + I  $$  Remembering (*) I get: $$  2.28319 = 0$$, which cannot be.   I am have run my calculations down from top to bottom and up again and seem not able to find the mistake. I would be very happy, if someone could point it point it out in the comments.
32523
I'm trying  to calculate $\int \frac{x^2 }{(x^2+1)^2} dx$ by using the formula: $$ \int udv = uv -\int vdu $$  I supposed that $u=x$ s.t $du=dx$, and also that $dv=\frac{x}{(x^2+1)^2}dx$, but I couldn't calculate the last integral. what is the tick here?           the answer must be: $ \frac{1}{2}arctan\ x - \frac{x}{2(1+x^2)} $ + C
52401
How to evaluate the integral $$\int_{25\pi / 4}^{53\pi / 4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx\ ?$$
109848
If $ \frac {x^2}{a^2} + \frac {y^2}{b^2} = r^2$ is an ellipse, with the parameterization $x(θ)≔r(a \cos ⁡θ,b \sin⁡ θ ),$ I have to find the value of $θ$ giving the minimum distance from $P(p,q)$ (not on the ellipse) to the ellipse is given by a quartic in $t= \tan⁡( \frac {θ}{2}).$ A necessary condition for $x$ to be the closest point to $P$ is that $P-x$ is perpendicular to the tangent vector in $x ,$ i.e. $(P-x(θ) ). x' (θ)=0$  I can't handle the above condition to make an fuction (e.g. $f(θ)$) to find the minimum value by calculating the rerivative $f'(θ)=0$ for example. Then I have to prove that rational, non-zero values of $a, b, p, q$ can be found such that the quartic factorises as the product of two quadratics with rational coefficients. Any help? Thank you
51644
I know that $\int{\frac{1}{x}}dx$ is simply $\ln{(x)}+c$ (-which is clearly unrelated to the problem but I just thought I would share anyway) but I am not sure how to approach $e^{x{^2}}$. Perhaps a substitution?
27394
Show that   $$\int_{0}^{\pi} xf(\sin(x))\text{d}x = \frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))\text{d}x$$  this is so confusing. i have no idea how to even start. im thinking integration by parts but that doesnt seem right.  hope you guys can help!!
17860
I've already see a  by Marko Riedel which I list it follows:     The standard way to treat these sums is to integrate   $$ f(z) = \frac{1}{(z+\alpha)^2} \pi \cot(\pi z)$$   along a contour consisting of a circle of radius $R$ and with $R$ going to infinity and hence being larger than $\alpha$, where the circle does not pass through the poles on the real axis.   Now along the semicircle in the upper half plane we have   $$|f(z)| \le \frac{1}{(R-|\alpha|)^2}\pi \left|\frac{e^{i\pi R\exp(i\theta)} + e^{-i\pi R\exp(i\theta)}} {e^{i\pi R\exp(i\theta)} - e^{-i\pi R\exp(i\theta)}}\right|= \frac{1}{(R-|\alpha|)^2} \pi  \left|\frac{e^{2i\pi R\exp(i\theta)}+1}{e^{2i\pi R\exp(i\theta)}-1}\right| &lt; \frac{1}{(R-|\alpha|)^2} \pi   \frac{1+e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}} {1-e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}}$$   This last term is clearly $O(1/R^2)$ as $R$ goes to infinity as the quotient of the two exponentials goes to one since $\exp(-R)$ vanishes and there is no singularity when $\theta = 0$ or $\theta = \pi$ as $R\cos\theta = \pm R$, which is not an integer by the assumption that the circle avoids the poles and hence cannot be one.     My question: It is clear that for each $z=Re^{i \theta}$ , $f(z)=O(1/R^2)$ ，but can this guarantee that the last term on RHS is uniform bounded ? Since $\theta$ varies on a compact interval $[0, \pi]$ , I want to show that for each $ \theta \in [0, \pi]$ , and a fixed positive number $M \gt 1 $ there exist an open ball contains $\theta$ and a fixed $R_{\theta}\gt 0$ , for every $R \ge R_{\theta}$ and $x \in $ the open ball $$\frac{1+e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} {1-e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} \le M$$  For $\theta \neq 0,\pi$ , it is easy to find the desired $R_{\theta}$ and the open ball , but if $\theta = 0$ or $ \theta = \pi$ , for every open ball containing $\theta$ , it behave erratically near $\theta$ , I have no idea how to deal with this .
96009
$$\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}$$  I tried substituting $x=z^2$ ... also $x=\tan^2 \theta$ ... but couldn't solve it either ways... if someone can help then it would be good.
45800
Plot $|z - i| + |z + i| = 16$ on the complex plane   Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellipse algebraically.  I get to the point:  $x^2 + (y - 1) ^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} + x^2 + (y+ 1)^2 = 256$  $2x^2 + 2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$  It seems like I'm and doing something the hard way.