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You can use a little bit of instinct for these things. Sure, you can just go and check the slope. But there are other signs as well. First of all, an integral of an odd function is even and vice-versa (if you disregard the constant offset). So, (A) and (B) are immediately out. Then, recall that integration smooths things. So integrating a discontinuous ...

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As $f$ is odd, $F$ must be even. This excludes (A) and (B). And $(D)$ is excluded because its derivative is zero in all the interval except in a finite number of points (where doesn't exists).

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Adding up the integrands so that all terms are over the common denominator $x$, we find that the resultant numerator factors in a very helpful way. As is, spotting a convenient way to group the terms isn't exactly easy, but if you're familiar with the Nielsen generalized polylogarithm you'll find that it tidies things up quite nicely. In particular, note ...

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As I already commented $$I=\int\frac{x}{\cos(x)}\,dx$$ looks to be a monster. What I am ready to bet is that the problem is $$J=\int\frac{x}{\cos^2(x)}\,dx$$ which is a totally different story. Integrating by parts $u=x$, $v'=\frac{dx}{\cos^2(x)}$, $u'=dx$, $v=\tan(x)$ give $$J=x \tan(x)-\int \tan(x)\,dx=x \tan(x)-\int \frac{\sin(x)}{\cos(x)}\,dx=x ... 0 You need to look at which of the functions A, B, C or D, when differentiated, gives f(x). Remember that when you differentiate, you find the value of the slope of the curve of the function you're differentiating. For instance, look at C: In the negatives, it's slope is negative, so it's derivative must yield values <0 in this interval. In the ... 0 Using the disc method, you have that V=\int_{-1}^1 π(R^2-r^2)dy where R=8-y^2 and r=8-1, so V=\int_{-1}^1 π((8-y^2)^2-7^2)dy 0 The area being rotated is (1;8)\times(0;\sqrt x) or (0;8)\times(0;1)\cup(1;y^2)\times(1;\sqrt 8) This is being rotated around the vertical line x=8, so either use 2\pi \int_1^8 r~y\operatorname d x or \pi(7^2+\int_1^{\sqrt 8} r^2\operatorname d y). Can you figure out what r is in terms of x and y respectively? 0 You can proceed the same way as you did before. Usually, the divergence theorem suffices to calculate the flux of a vector field through a surface in \mathbb{R}^n, and is stated as: In a compact V\subset \mathbb{R}^n with a piecewise smooth boundary \partial V and a vector field \vec F defined in a boundary of V, \int_{\partial V} \vec F \cdot ... 1 You are rotating about a vertical line. So you can use the "discs" or "washers" method, integrating with respect to y; or the "cylindrical shells" method, integrating with respect to x. But you are trying to use the "discs" method and integrating with respect to x. Hint: draw a picture! 0 Disclaimer: I know this isn't really a proper answer, but I couldn't post the picture I took in the comments, so I hope it's okay. I found a section in his Quantum Mechanics book that goes through a very similar calculation, which should also help you with the n = n' case. The functions \psi(x) are just some sines in this case (as you can see) - you can ... 0 The solution appears to be correct. Personally, I used different construction of the integral, which is \int_0^2dy\int_0^{2-y}dz\int_0^{4-y^2}dx = \frac{20}{3}. Hope this helps. 0 Sketching is indeed a good idea when you deal with non-trivial boundaries. In this case, the main boundaries are the coordinate planes x=0, y=0, and z=0. Then, by inspection, we notice that the given plane x = 2- 2y makes the bound region (in x-y plane) in the first octant, which means that our boundaries become x\in[0, 2-2y], y\in [0, 1] ... 0$$\int\frac{x^7}{\left(1-x^4\right)^2}\space\text{d}x=\int\frac{x^7}{\left(x^4-1\right)^2}\space\text{d}x=$$Substitute u=x^4 and \text{d}u=4x^3\space\text{d}x:$$\frac{1}{4}\int\frac{u}{\left(u-1\right)^2}\space\text{d}u=\frac{1}{4}\int\left[\frac{1}{u-1}+\frac{1}{(u-1)^2}\right]\space\text{d}u=$$... 4 HINT:$$\dfrac14\int\dfrac{x^4}{(1-x^4)^2}\cdot4x^3\ dx$$Set 1-x^4=u 3 I will use I(k) for the integral instead of f(k).$$ I(k)=\int^1_0 t^2 \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}}dt $$First, let's find some particular value, we will need it later.$$I(1)=\frac{1}{3} $$Now the definition for the elliptic integral of the second kind:$$ E(k)=\int^1_0 \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}}dt $$It's easy to show that: ... 1 You are almost done! Use$$\tan(\arctan(x))=x$$(per definition) and$$\sec(\arctan(x))=\sqrt{x^2+1}$$(convince yourself of this by looking at a right-angled triangle for a few seconds) to arrive at the result$$\pi\sqrt{4\pi^2+1}+\frac12\ln\left(\sqrt{4\pi^2+1}+2\pi\right)$$There is probably no easy way to approximate that number without a ... 1 It ultimately relies on the two observations$$\underbrace{c+c+\cdots+c}_{n\textrm{ times}} =n\cdot c$$and$$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$You should memorize these formulas, they are used rather frequently. Note that the summation notation disguises them somewhat. They could be written as$$\sum_{k=1}^n c =n\cdot c\sum_{k=1}^n k ...

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For the first series: $$\frac{12}{n}\sum_{i=1}^n -7=\frac{12}{n}\sum_{i=1}^n -7\cdot1=\frac{12}{n}\cdot(-7)\sum_{i=1}^n 1=\frac{-84}{n}\sum_{i=1}^n 1$$ Similar for the second one. Then, use the fact that $$\sum_{i=a}^b 1=(b-a+1)$$ and $$\sum_{i=a}^b i=\frac{(a+b)(b-a+1)}{2}$$ for $a=1$ and $b=n$.

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Hint To compute $$I=\int \sec^3(x)\,dx$$ the tangent half-angle substitution seems to be convenient. $$t=\tan(\frac x2) \qquad dx=\frac{2\, dt}{1+t^2}\qquad \sec(x)=\frac{1+t^2}{1-t^2}$$ make $$I=2\int\frac{ \left(1+t^2\right)^2}{\left(1-t^2\right)^3}\,dt$$ Now, partial fraction decomposition leads to very simple terms.

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After this corps has rised from the dead anyway, let me give an additional solution which is based on complex analysis but avoids the use of branch cuts, so it should be viewed as complementary to @RandomVariables approach. First, perform an subsitution $x\rightarrow\sin(t)$ which brings our integral into the form $$... 0 You should recognize the derivative of the cosecant. Then by parts,$$\int \frac{x\cos x}{\sin^2x}dx=-\frac x{\sin(x)}+\int\frac{dx}{\sin(x)}.$$The last integral is easily solved with$$\int\frac{dx}{\sin(x)}=\int\frac{\sin(x)}{\sin^2(x)}dx=\int\frac{\sin(x)}{1-\cos^2(x)}dx=-\text{artanh}(\cos(x)).$$2 Notice, using integration by parts$$\int \frac{x\cos x}{\sin^2 x}\ dx=x\int \frac{\cos x}{\sin^2 x}\ dx-\int \left(\frac{d}{dx}(x)\cdot \int \frac{\cos x}{\sin^2 x}\ dx\right)\ dx=x\int \frac{d(\sin x)}{\sin^2 x}-\int \left(1\cdot \int \frac{d(\sin x)}{\sin^2 x}\ dx\right)\ dx=x\cdot \frac{-1}{\sin x}-\int \left(\frac{-1}{\sin x}\right)\ ...

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Since you wanted to solve using partial fractions, $$\int\frac{x\cos x}{(1+\cos x)(1-\cos x)}dx=\frac{1}{2}\int\frac{x(1+\cos x)-x(1-\cos x)}{(1+\cos x)(1-\cos x)}dx$$ $$=\frac{1}{2}\int\frac{x}{1-\cos x}dx- \frac{1}{2}\int{\frac{x}{1+\cos x}}dx$$ $$= \frac{1}{2}\int \frac{x}{2\sin^2\frac{x}{2}}dx- \frac{1}{2}\int \frac{x}{2\cos^2\frac{x}{2}}dx$$ $$= ... 2 For the calculation of$$I=\int \csc(x)\,dx$$the tangent half angle subsitution is very useful and makes everything simple$$t=\tan(\frac x2)\implies dx=\frac {2dt}{1+t^2}\qquad \csc(x)=\frac{1+t^2}{2t}$$All of this makes$$I=\int \frac{dt} t$$1$$\int \frac{x\cos x}{(1-\cos x)(1+\cos x)}dx=\int\frac{x \cos x}{1-\cos^2 x}dx=\int x\cot x \csc xdx$$And now by parts: f=x and df=dx and dg=\cot x \csc x dx and g=-\csc$$=-x \csc x+\int\csc x dx$$Multiply numenator and denominator of \csc x by \cot x +\csc x$$-x\csc x-\int\frac{-\cot x \csc x-\csc^2 x}{\cot x+\csc x}dx$$And now set ... 1$$\int \frac{x\cos x}{\sin^2x}dx$$Let \sin x=u. Then, \cos x dx=du and x=\arcsin u.$$\int \frac{x\cos x}{\sin^2x}dx=\int \frac{\arcsin u}{u^2}du$$This can be integrated by parts taking \arcsin u as the first function and \frac{1}{u^2} as the second function. 2 Note that$$\frac{\cos x}{\sin^2 x}=\csc x \cot x$$Applying integration by parts and differentiating 'x' we get,$$I= \int \frac{x\cos x}{\sin^2x}dx=-x \csc x +\int \csc x\,dx$$Now can you proceed? 0 \displaystyle J=\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx Perform the change of variable y=\sqrt{\dfrac{1-x}{1+x}}, \displaystyle J=-4\int^1_0 \dfrac{\log(x)}{1-x^2}dx=-4\int_0^1 \left( \log x\times\sum_{n=0}^{+\infty}x^{2n}\right) dx=-4\sum_{n=0}^{+\infty}\left(\int_0^1 x^{2n}\log x dx\right) \displaystyle ... 0 We have f(x) = f(1)+\int_1^x f'(t)dt. By assumption the limit for x \to \infty of the right hand side exists, hence \lim\limits_{x \to \infty} f(x) =: c exists. It is pretty obvious that \int_{1}^\infty |f(x)|dx < \infty implies c=0. You should work out at least this part by yourself. -1$$-\alpha x^2+\beta x=-\alpha\left(x^2-\frac\beta\alpha x\right)=-\alpha\left(x-\frac\beta{2\alpha}\right)^2+\frac{\beta^2}{4\alpha}\impliesI(\alpha):=\int_{\Bbb R}e^{-\alpha\left(x-\frac\beta{2\alpha}\right)^2}e^{\frac{\beta^2}{4\alpha}}\;dx=e^{\frac{\beta^2}{4\alpha}}\sqrt{\frac\pi{\alpha}}\implies\implies I'(\alpha)=\sqrt\pi\; ...

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The orthogonal complement of $Y$ consists of all $g$ such that $$0 = (f,g) = \int_{-\pi}^{0}f(t)\overline{g(t)}+\int_{0}^{\pi}f(t)\overline{g(t)}dt \\ = \int_{0}^{\pi}f(t-\pi)\overline{g(t-\pi)}+f(t)\overline{g(t)}dt \\ = \int_{0}^{\pi}f(t)\overline{\{g(t-\pi)+g(t)\}}dt,\;\;\; f \in Y.$$ It follows that $$Y^{\perp} = \{ ... 0 I assume that you are required to interchange the order of integration. Because the order you've selected is much easier. In the opposite order, the region is defined by: for -1 \leq y \leq -\frac{1}{\sqrt{2}}, the inequalities \arccos y \leq x \leq 2\pi - \arccos y; for -\frac{1}{\sqrt{2}} \leq y \leq \frac{1}{\sqrt{2}}, the inequalities \arccos y ... 2 Start with evaluating the integral$$I=-\int_{R} e^{-\alpha x^2+\beta x} dx Which is Gausian type then differentiate the answer with respect to alpha. Note: you can use the new formula I derived a while ago which handles a more general integral than yours. See here. 1 I agree entirely with user8268's comment. You don't need to worry about this being an approach to integration "for dummies." Apostol's book is completely rigorous, except for a handful of places in the second volume where a difficult proof may be omitted, and even when that happens he is always completely upfront about it. If you are interested in ... 0 W|A can integrate that and show the process of derivation. You can also check and verify the answer you arrived at (|\cdot| omitted) evaluates to the integrand when differentiated, which is a sufficient proof of validity by definition of integral. On your actual question, I am fairly sure the issue here is exactly the same one as with the infamous ... 0 Note that \frac{1}{x} is an odd function, then \int \frac{dx}{x} is even whatever the integration constant is. Therefore we can make the anti-derivative even by inserting the absolute sign, that is \int \frac{dx}{x}=\ln |x|+C. This can be verified by differentiate the RHS: for x<0, (\ln |x|)'=[\ln (-x)]'=-\frac{1}{(-x)}. Similar reason for ... 0 Consider the addition case. First use x = a \text{ csch} u  and dx = -a \text{ csch} u \text{ coth} u du \begin{align} \int \frac{dx}{x\sqrt{a^2 + x^2}} &= \int \frac{-a (\text{ csch} u) (\text{ coth} u) du}{(a \text{ csch} u)(a \text{ coth} u)}\\ &= -\int \frac{1}{a} du\\ &= -\frac{1}{a}u + c\\ &= - \frac{1}{a}\text{ arcsch} \frac{x}{a} ... 1 The substitution x = \tan \theta leads to \int_{\arctan a}^{\arctan b} \cos^{-2s-2} \theta \, d\theta. According to Wolfram Alpha, this can be integrated using a hypergeometric function. Namely,\int (1 + x^2)^s \, dx = x\,{}_2F_1(1/2, -s, 3/2, -x^2) + C,$$where {}_2F_1 is the function described here. 0 See this answer for a proof that for f\in \mathscr R[a,b] and given any \epsilon>0, one can find g\in\mathscr C[a,b] such that$$\int |f-g|<\epsilon. $$Now note that$$|\int f^2|\le|\int (f^2-fg)|\le \int |f^2-fg|\le \sup |f|\int |f-g|\le \sup |f|\epsilon. $$By our arbitrary choice of \epsilon it follows that$$\int f^2=0. $$Take it from ... 2 Let g_N(\theta_N) be given by$$g_N(\theta_N)=\int_0^{\theta_N}\frac{\sin\left((N+1/2)x\right)}{\sin(x/2)}\,dx-\pi$$where \theta_N=\pi/(N+1/2). Now, enforcing the substitution x\to x/(N+1/2) yields$$\begin{align} g_N(\theta_N)&=\int_0^{\pi}\frac{\sin(x)}{\sin\left(\frac{x}{2N+1}\right)}\frac{1}{N+1/2}\,dx-\pi\\\\ ...

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In a comment, you wondered if the problem could be instead $$\int \frac {\frac 12-u^2}{2u^4-2u^2-1}\, du$$ Doing the same as before, we have $$2x^2-2x-1=2(x-a)(x-b)$$ with $$\quad a=\frac{1}{2} \left(1-\sqrt{3}\right)<0\quad,\quad b=\frac{1}{2} \left(1+\sqrt{3}\right)>0$$ and, again, partial fraction decomposition makes$$\frac{1-2 x}{2 \left(2 x^2-2 ... 1 Hint: Show that if f is not 0 at a continuity point, then f(x) > 0 or f(x) < 0 on some interval [c_1,c_2]. Now choose$$h(x) = \begin{cases} (x-c_1)(c_2-x), \,\, &x \in [c_1,c_2],\\ 0, \,\, &\mbox{otherwise}\end{cases}$$5 OP, you are correct,$$\int_1^n\frac{\sin \sqrt{x}}{\sqrt{x}}dx=-2\cos\sqrt{n}+2\cos 1$$Hence the improper integral is$$\int_1^\infty\frac{\sin \sqrt{x}}{\sqrt{x}}dx=\lim_{n\to\infty}\int_1^n\frac{\sin \sqrt{x}}{\sqrt{x}}dx=\lim_{n\to\infty}-2\cos\sqrt{n}+2\cos 1$$And the latter limit does not exist. Your computer algebra system (mathematica) is giving ... 0 If you differentiate \ln x, then you get \frac{1}{x}\;(x>0), and if you differentiate \ln (-x), then you get \frac{1}{x}\;(x<0). Therefore, if you find the indefinite integral of \frac{1}{x}, then$$ \int \frac{1}{x} dx = \frac{1}{|x|}+C. $$Using partial fraction,$$ ...

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Observe first that $$\int_{-\infty}^{+\infty}\sin(k_0\xi)e^{-\frac{(x-\xi)^2}{4a^2t}}\,d\xi =\Im\left[\int_{-\infty}^{+\infty}e^{ik_0\xi}e^{-\frac{(x-\xi)^2}{4a^2t}}\,d\xi\right]$$ Then notice that \begin{align*} ik_0\xi-\frac{(x-\xi)^2}{4a^2t} &=-\frac1{4a^2t}\left[\xi^2-2\left(ik_02a^2t+x\right)\xi+x^2\right]\\ ...

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Kim Peek's "funny" hypergeometric solution is really the series solution near $x=0$. We have, for $|x^a/b| < 1$, $$\sqrt{x^a + b} = \sqrt{b} \sqrt{1 + x^a/b} = \sqrt{b} \sum_{k=0}^\infty {1/2 \choose k} (x^a/b)^k$$ so integrating term-by-term $$\int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{x^{ak+1}}{(ak+1) \; b^{k-1/2}}$$ ...

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Lots of integrals of that kind cannot be solved in terms of elementary simple functions. Indeed, there is a large class of integrals whose solutions are the so called Jacobi Elliptic Integrals Functions (of the first kind, second kind and third kind). Generally, you have an integral of the form $$\int R\left(x, \sqrt{P(x)}\right)\ \text{d}x$$ where $P(x)$ ...

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Your own approach was easier than some of those suggested. Picking up from where you left off, you just use the double angle formula and reversing the chain rule, and we have $$\frac 14\int \frac 12(1-\cos 4x)dx-\frac 14\times \frac 16\sin^32x$$ Can you finish this now?

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Just expand term-by-term : $$(f*g)(t)=\int_0^6f(x)g(t-x)dx$$ $$=2\int_0^6[u(x)u(t-x)-u(x-5)u(t-x)-u(x)u(t-x-1)+u(x-5)u(t-x-1)]$$ $$=2(u(x)*u(x)-u(x-5)*u(x)-u(x)*u(x-1)+u(x-5)*u(x-1))$$ $$=2(tu(t)-(t-5)u(t-5)-(t-1)u(t-1)+(t-6)u(t-6))$$ Actually we don't have to expand the terms since convolution is linear operation and I used $u(x-a)*u(x-b)=(t-a-b)u(t-a-b)$. ...

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$$\cos^2x = 1 - \sin^2x$$ Known this and the game is done. Then you have to compute $$\int\sin^4(x) - \sin^6(x)\ \text{d}x$$ Which is quite easy. Can you proceed? Hint Reduction formula $$\int\sin^m(x)\ \text{d}x = -\frac{\cos(x) \sin^{n-1}(x)}{n} + \frac{n-1}{n}\int \sin^{n-2}(x)\text{d}x$$

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