I'm trying to find the mass of a spherical object with a given density function, and to do so I must solve this integral $$\int_{0}^{\pi}\sin^5{\theta}\cos^2{\theta}\ d\theta,$$  but no matter which method I choose (integration by parts, substitution, etc) I can't for the life of my figure out the anti-derivative.
15156
Suppose $\displaystyle P(x,y)$ a polynomial in the variables $x,y$.   For example, $\displaystyle x^4$ or $\displaystyle x^3y^2 + 3xy + 1$.  Is there a general method which allows us to evaluate the indefinite integral     $$ \int P(\sin x, \cos x) \text{d} x$$   What about the case when $\displaystyle P(x,y)$ is a rational function (i.e. a ratio of two polynomials)?   Example of a rational function: $\displaystyle \frac{x^2y + y^3}{x+y}$.    This is being asked in an effort to cut down on duplicates, see here:   and here:
37731
I'm not entirely sure what I want to ask here, so please bear with me!  I think the explanation they give us in school for how finding the antiderivative by substitution works is: $$\int f(g(t))g&#39;(t)dt=(Fg)(t)=F(g(t))=F(u)=\int f(u)du$$  But I never really understood the equality $F(u)=\int f(u)du$. Why can we behave as if the identity $u=g(t)$ doesn't exist, compute the integral $\int f(u)du$, and then 'substitute' $u$ in the result yielding the correct answer? In a sense, it feels like the variable $u$ switches from being 'meaningful' in $F(u)$ (where it stands for $g(t)$) and being insignificant in $\int f(u)du$, since as I perceive it the "variable name" here is meaningless and we could say $ \int f(u)du = \int f(k)dk = \int f(x)dx = ... $ all the same. Another way to ask the same thing: why $(Fg)(t)=\int f(u)du$ and not, say, $(Fg)(t)=\int f(x)dx$? Where am I getting confused?  Maybe someone can explain this better than my high school teacher? Is there a 'formal' explanation for this? Thank you a lot!
22559
Okay I like have no idea how to do this. I tried integration by parts then everything came back uglier. I then put this in wolfram it says "No found in standard mathematical functions"   Is there a easy way to do this?   Maybe like FTOC II?
113522
Please refer to my school textbook pg48 (of the book, and not the pdf counter) here:   My doubt is in this context: (right side column)  $a = dv/dt = v dv/dx$  then integrating $v$ with respect to $dv$, and $a$ with respect to $dx$.  Now, when we integrate $a$, either we say it is a constant, and give $a\times(x-x_o)$ as a result of the integration, or we say it is not constant, and is a function of time, in which case, it can not be integrated like this. But the book integrates it like this, and then it is written:     The advantage of this method is that it can be used for motion with non-uniform acceleration also.   Can you please explain how it can be used with non uniform acceleration while it has been assumed while integrating that $a$ is constant? And moreover, does this line seem to apply to only this derivation, or does it apply to other two before it as well?
116928
I was experimenting with various integration techniques, and I stumbled upon the integral:  $$ \int{\frac{\tan{x}}{x} dx} $$  I tried using a number of methods, but I could not solve it. Checking on WolframAlpha, I found that the integral does not have an elementary antiderivative. However, I am not satisfied with simply assuming it's non-elementary if I have difficulty in solving it.   My question: Is there a test to determine if an antiderivative is elementary or not?
46433
I'm asked to show  $\int_0^{\infty} \frac{x\sin ax}{x^2+t^2}dx = \frac{\pi}{2}e^{-at}$ when $t,a &gt; 0$. I tried using integration by parts integrating $\frac{x}{x^2+t^2}$. But it seems that $\frac{x\sin ax}{x^2+t^2}$ has no antiderivative. What should I try?
42871
So one of the question on the MIT Integration bee has baffled me all day today $$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$$ I have tried a variety of things to do this, starting with Integration By Parts Part 1 $$\frac{\tan x-x}{1+x^2}\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\frac{-2x(\tan x -x)}{\left (1+x^2 \right )^2}\text{d}x$$ which that second integral is not promising, so then we try Integration By Parts Part 2 $$\tan^{-1} x\tan^2 x\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}2\tan^{-1} x\tan x\sec^2 x\text{d}x$$ which also does not seem promising Trig Substitution $x=\tan\theta$ which results $$\int_{0}^{\tan^{-1}\frac{\pi}{4}}\tan^2 \left (\tan\theta\right )\text{d}\theta$$ which I think too simple to do anything with (which may or may not be a valid reason for stopping here) I had some ideas following this like power reducing $\tan^2 x=\frac{1-\cos 2x}{1+\cos 2x}$ which didn't spawn any new ideas. Then I thought maybe something could be done with differentiation under the integral but I could not figure out how to incorporate that. I also considered something with symmetry somehow which availed no results. I'm also fairly certain no indefinite integral exists. Now the answer MIT gave was $\frac{1}{3}$ but wolfram alpha gave $\approx$ .156503. Note  The integral I gave was a simplified version of the original here is the original in case someone can do something with it $$\int_{0}^{\frac{\pi}{4}}\frac{1-x^2+x^4-x^6...}{\cos^2 x+\cos^4 x+\cos^6 x...}\text{d}x$$ My simplification is verifiably correct, I'd prefer no complex analysis and this is from this  close .
28856
I understand how derivatives work based on the definition, and the fact that my professor explained it step by step until the point where I can derive it myself.  However when it comes to the area under a curve for some reason when you break it up into an infinite amount of rectangles, magically it turns into the anti-derivative. Can someone explain why that is the definition of the integral and how Newton figured this out?
84718
I am trying to establish that  $$\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2\left(\frac{\theta}{2}\right)\;\mathrm d\theta=2f(0)+f^\prime(0)$$  for $f$ analytic on $|z| &lt; R,\; R &gt; 1$  I think I am making a mistake somewhere, please point it out to me.Here's what I've done unsing the mean value property and Cauchy's integral formula:  $$2f(0)+f&#39;(0)=2(\frac1{2\pi}\int_0^{2\pi} f(e^{i\theta})d\theta + \frac1{2\pi i}\int_0^{2\pi}\frac{f(e^{i\theta})ie^{i\theta}}{e^{i2\theta}}=\frac{1}{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta\frac1{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}}\quad\text{(1)}$$  But when I develop the expression $\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2(\frac{\theta}{2}) d\theta$ I get:  $$\frac1{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta+\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})[{e^{i\theta}}+e^{i-\theta}]d\theta \quad\text{(2)}$$  The second integral in (1) differs from the second integral in (2). Thanks.
101742
I need solve this integral, and I tried various methods of solving and did not get it. The integral is:   $$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta,$$  where $t$ is a positive integer.
74912
I would like to show $\lim\limits_{r\to\infty}\int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$.  Now, of course, the integrand does not converge uniformly to $0$ on $\theta\in [0, \pi/2]$, since it has value $1$ at $\theta =0$ for all $r\in \mathbb{R}$.   If $F(r) = \int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta$, we can find the $j$th derivative $F^{(j)}(r) = (-1)^j\int_{0}^{\pi/2}\sin^{j}(\theta)e^{-r\sin\theta}\text d\theta$, but I don't see how this is helping.  The function is strictly decreasing on $[0,\pi/2]$, since $\partial_{\theta}(e^{-r\sin\theta})=-r\cos\theta e^{-r\sin \theta}$, which is strictly negative on $(0,\pi/2)$.  Any ideas?
47523
Evaluate $$\int_{0}^{2\pi}\frac{1}{(1+a\cos {\theta})^2}\,d\theta \quad , \, 0\le a&lt;1$$  I know it can be solved using complex analysis, but how do I solve this with real analysis methods? This integral appears in the general derivation of Kepler's third law. Here $a$ is the eccentricity of the planetary orbit. So $a=0$ was assumed to prove that $T^2\propto r_0^3$ for a circular orbit, where $T$ is the orbital period and $r_0$ is the orbital radius.
109003
What is the value of $$\int_0^{\pi} e^{a\cos(t)}\cos(a\sin t)dt?$$
122862
I need to solve $\int \frac{\sqrt{25-y^2}}{y}dy$.  I originally thought IBP, but that led to a very large and confusing algebra problem.   Then I started to look at the $\sqrt{25-y^2}$ and started to consider doing a $u$-substituion, and letting $u=\sin\theta$ and $du=\cos\theta d\theta$. Once I started playing a bit, the intuition was lost.  Any and all help is appreciated.
50265
$F$ is a distribution function .Take any $a \in \mathbb{R}$.  Then evaluate : $\displaystyle\int_{\mathbb{R}} (F(x+a)-F(x)) dx$   As it is a general distribution function , I cannot employ densities to solve this problem . Any insights or a complete solution will be very appreciated .
47738
While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.      $$\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.$$
57798
$\int e^{2\theta}\ \sin 3\theta\ d\theta$  After Integrating by parts a second time, It seems that the problem will repeat for ever. Am I doing something wrong. I would love for someone to show me using the method I am using in a clean and clear fashion. Thanks.