change the order of integration in $$\int_0^{2a}\int_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)x}{\sqrt{4a^2x^2-(x^2+y^2)^2}}dy\,dx$$  I was able to change the order of integration here to  $$\int_0^a\int_{a-\sqrt{a^2-y^2}}^{a+\sqrt{a^2-y^2}}\frac{\phi'(y)(x^2+y^2)x}{\sqrt{4a^2x^2-(x^2+y^2)^2}}dx\,dy$$ Now i am stuck with the integration here w.r.t $x$. I tried substituting $(x^2+y^2)^2$ with $t$ but then replacing value of $4a^2x^2$ becomes a problem. I was thinking of substituting both $x$ and $y$ as $k\cos\theta$ and $k\sin\theta$ but then $y$ won't be constant w.r.t $x$, so that too I believe is out of the question!
32812
Change the order of integration in:  $$\int_0^{2a}\int_0^\sqrt{2ax-x^2}\frac{\phi'(y)(x^2+y^2)x }{\sqrt{4a^2x^2-(x^2+y^2)^2}}dxdy$$ and hence evaluate it.    I changed the order of integration and got the limits $y=0$ to $a$ and $x=a-\sqrt{a^2-y^2}$ to $a+\sqrt{a^2-y^2}$. I then tried changing to polar coordinates, but it is only making the integrand more complex.
94094
$$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$ How to integrate? I tried the substitution $x=\sin\theta$ but didn't work.
29505
$$ \int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x $$   I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and $(x)^2-1/(x)^2=z $ but no helpful expression was derived. I also used property $\int_0^a f(a-x)=\int_0^a f(x) $ Please help me out
72851
Could someone help with the following integration:   $$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$  So far I have done the following, but I am stuck:  I denoted $ y=-\cos x $ then: $$\begin{align*}&amp;\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&amp;= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&amp;=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$  Then I am really stuck. Could someone help me?
93963
I don't know how to integrate $\int x^2\sin^2(x)\,\mathrm dx$. I know that I should do it by parts, where I have  $$ u=x^2\quad  v'=\sin^2x \\  u'=2x \quad v={-\sin x\cos x+x\over 2}$$  and now I have  $$ \int x^2\sin^2(x)\,\mathrm dx = {-\sin x\cos x+x\over 2} x^2 - \int 2x{-\sin x\cos x+x\over 2}\,\mathrm dx\\ ={-\sin x\cos x+x\over 2} x^2 - \int x(-\sin x\cos x+x)\,\mathrm dx\\  $$  so I have to calculate $$ \int x(-\sin x\cos x+x)\,\mathrm dx=-\int x\sin x\cos x\,\mathrm dx+\int x^2\,\mathrm dx$$  I know that $\int x^2\,\mathrm dx = {1 \over 3}x^3+C$ but I don't know what should I do with $$\int x\sin x\cos x \,\mathrm dx$$ Should i use parts again or what? Please help.
106957
$$\int_0^\pi x \ln(\sin (x))dx  $$   I tried integrating this by parts but I end up getting integral that doesn't converge, which is this $$ \int_0^\pi \dfrac{x^2\cos (x)}{\sin(x)} \ dx$$ So can anyone help me on this one?
66145
I have a calculus II final coming up and this question came up in a past final exam:  $$\int_{6}^{16} \frac{dx}{\sqrt{x^3 + 7x^2 + 8x - 16}} = \frac{\pi}{k},$$  where $k$ is a constant. Find $k$.  My progress so far:  $$\int_{6}^{16} \frac{dx}{\sqrt{(x - 1)}(x + 4)} = \frac{\pi}{k}$$  The answer is: $k = 6\sqrt{5}$  I do not know where to from from this step. Any helps or hints will be greatly appreciated! Thank you.  EDIT: $(x + 4)$ not $(x + 2)$ in the denominator.
95302
Solve the DE $$yy^{\prime}+x=\sqrt{x^2+y^2},x&gt;0$$ I show it is a homogenous first order DE and I use the substitution $y=Vx$. Then, I solve until $$\int \frac{V}{\sqrt{1+V^2}-(1+V^2)}dV=\int \frac{1}{x}dx$$ I don know how to integrate LHS. Any hint?
88098
$\int e^{\frac{x^2}{2}} \cdot x^{3} dx=?$  I tried to do the substitution $du = x^3dx$, so that $u=\frac{x^4}{4}$. Then, $\int e^{\frac{x^2}{2}} \cdot x^{3} dx= \int e^{\sqrt{u}} du$  However, I would have to do a second substitution ($v = \sqrt{u}$) which would give a product again.  How should I proceed?
117001
Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$ What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$
57872
How can I rewrite, say: $$\int^1_0\int^{z^2}_0\int^y_0{f}dxdydz$$ In the orders $dzdxdy$ and $dydzdx$?  Would I have to solve this triple integral all the way or is there a simple way to do this?
109306
So I have recently learnt to calculate &quot;Double Integrals&quot; and I have a doubt on change of variables. Lets say we are evaluating a double integral of the form I = $\int\int _R F(x,y) dx dy $ , where $F(x,y)$ is an integrable smooth function and $R$ is the region of integration in the $xy$ plane. Now I want to convert to polar co-ordinates $x(r,\theta)=r \cos(\theta)$ and $y(r,\theta) = r \sin(\theta)$. Now the region $R$ in the $xy$ palne can be suitably converted to some region $R'$ in the $r \theta$ plane. The function $F(x,y)$ may take the form of some smooth and integrable $G(r,\theta)$. My main doubt is with the change of the area element. According to my textbook, $dx dy = \frac{\partial (x,y)}{\partial(r,\theta)} = rdrd\theta$. So the area element in the $r\theta$ plane is $rdr d\theta$. Now I want to experiment a little bit with my understanding : I write $dx = d(r\cos\theta) = -r\sin\theta d\theta + \cos \theta dr$  and $dy = d(r\sin\theta)=r\cos\theta d\theta + \sin\theta dr $. And now I multiply the two expression for $dx$ and $dy$ : $dxdy= -r^2\sin\theta\cos\theta d\theta d\theta - r \sin^2 \theta drd\theta + r \cos^2 \theta dr d\theta + \sin \theta \cos\theta dr dr$. Why isn't this expression equal to $rdrd\theta$ ? The $d\theta d\theta$ and $dr dr$ terms look very odd to me. Am I commiting some abuse of notation or something ? Also there are no higher order terms like $dr dr d\theta$ which I can ignore. I am totally confused with all this. Can someone explain this to me a bit clearly? Thanks in Advance :). EDIT 1 : I know differentials are not the &quot;usual&quot; kind of algebraic quantities and we can't just add and multiply them casually. I am just multiplying $dx$ with $dy$ with the spirit of multiplying some $\delta x$ and $\delta y$ and letting both $\delta x$ and $\delta y$ tend to zero. EDIT 2 : Is there a branch of Mathematics where addition and multiplication of differentials is very common? Is there something like differential of a differential ?
61500
What I've done is this:$$\int\dfrac{\cos5x+\cos4x}{1-2\cos3x}{dx}$$ $$\int \dfrac{\sin 3x}{\sin 3x}\left[\dfrac{\cos5x+\cos4x}{1-2\cos3x}\right]{dx}$$ $$\dfrac {1}{2}\int\dfrac{\sin 8x -\sin 2x +\sin 7x -\sin x}{\sin 3x - \sin 6x}$$ $$-\dfrac {1}{2}\int\dfrac{ \sin \frac{7x}{2} +\sin \frac{5x}{2} } {\sin \frac{3x}{2} }$$ $$-\int\dfrac{ \sin {3x}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$ $$-\int\dfrac{2\sin \frac{3x}{2} \cos \frac{3x}{2}\cos \frac{x}{2} } {\sin \frac{3x}{2} }$$ $$-\int {2\cos \frac{3x}{2}\cos \frac{x}{2} }$$ $$ -\left(\frac{\sin 2x}{2} +\sin x \right) +c $$ Is there any other way to do so ? Is it possible to do it by substitution ?
77542
Find the integral  $$\int\!\sqrt{\cot x}\,dx $$ How can one solve this using substitution?  Can this be solved by complex methods?
60553
Calculate the following integral by hands: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin^{3}(t)}{\sin^{3}(t)+\cos^{3}(t)}dt$$  It seems apparently that integration by part does not work. I think integration by substitution could work but I can not figure out how to substitute.  I have calculated it by Wolfram and the result is $\frac{\pi}{4}$.
108659
How do I compute the integration for $a&gt;0$,   $$ \int_0^\pi \frac{x\sin x}{1-2a\cos x+a^2}dx? $$  I want to find a complex function and integrate by the .
105081
What I have done so far: Substituting $$x=3\sin(t)\Rightarrow dx=3\cos(t)dt$$ converting our integral to  $$I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\ \Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^2(t)\cos(t)}{3}+\frac{2}{3}\int\sin(t)dt\\ \Rightarrow I=18\cos(t)-9\sin^2(t)\cos(t)+c$$ Which is where I'm stuck. I can substitute back in for the $\sin(t)$ terms but don't know hot to deal with the $\cos(t)$ terms and back out information about $x$.   edited: because I miswrote the problem, I am sorry.