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6

Lets consider the indefinite integral, $$ \int \sec(x)\tan(x) dx = \int \frac{\sin(x)}{\cos^2(x)} dx$$ We can then perform a $u$ substitution with $u=\cos(x)$ and $du = -\sin(x) dx $ obtaining, $$ \int \sec(x)\tan(x) dx = -\int \frac{1}{u^2} du= \frac{1}{u} + C = \frac{1}{\cos(x)} +C = \sec(x) + C$$


6

Hint: Here's something for you to think about. By the Chain Rule $$\frac{d}{dx} \left(\cos (x)\right)^{-1} = (-1) (\cos (x))^{-2} \dot \, (-\sin (x)) = \frac{1}{\cos x} \dot\, \frac{\sin x}{\cos x}$$ Edit: In order to find $\int \sec x \tan x dx $. Write it as $$\frac{\sin x}{\cos^2 x}$$ and make the substitution $u = \cos x$.


5

To integrate a rational function there are two standard methods: the partial fractions decomposition, and the Ostrogradski-Hermite method. The latter does not require finding the roots of the denominator. See example here.


5

The first step is generally to divide out the integrand so as to get a polynomial plus a rational function whose numerator has lower degree than its denominator. Here you get $$\frac{x^3+2x}{1+x^2}=x+\frac{x}{x^2+1}\;.$$ Integrating the $x$ term (and, more generally, the polynomial quotient) is easy, so we’ve reduced the problem to integrating something of ...


3

$a)$ One thing you need to know by heart is this transformation: $x^2+y^2=r^2$, so the integrand becomes $\ln (1+r^2)$. We also have $dx\,dy=rdr\,dt$ (I will use $t$ instead of $\theta$ because it is easier for me) To compute the integral in the unit circle, you need to consider not $r=1$ but $r<1$ and $0<t<2\pi$ (they should have taught you this, ...


2

Let $\varepsilon>0$, then since $\int_{\Omega}|X|dP<\infty$, there exists an $n>0$ such that: $$nP(|X| > n) = \int_{\{\omega\in\Omega : |X(\omega)|>n\}}ndP \leq \int_{\{\omega\in\Omega : |X(\omega)|>n\}}|X(\omega)|dP < \varepsilon.$$ Therefore we have $$nP(|X| > n) < \varepsilon.$$


2

$$\int_\Omega |X|\, dP < \infty \implies \int_{\{|X| > n\}} |X|\, dP \to 0$$ by the dominated convergence theorem. Since $nP(|X|>n)$ is bounded above by the last integral, we have it.


2

Generally, you just try the long division if degree of $a(x)$ is higher than degree of $b(x)$. And for this particular example, write $\dfrac{x^3+2x}{1+x^2} = x + \dfrac{x}{1+x^2}$. Can you take it from here?


2

The expression $\sec x\tan x$ can be written $$ \frac{1}{\cos x}\frac{\sin x}{\cos x}=\frac{\sin x}{\cos^2 x} =-\frac{-\sin x}{\cos^2 x}=-\frac{f'(x)}{f(x)^2} $$ where $f(x)=\cos x$. Consider, for a generic differentiable function $f$, $$ g(x)=\frac{1}{f(x)}. $$ By the chain rule $$ g'(x)=-\frac{f'(x)}{f(x)^2}. $$ In the special case of $f(x)=\cos x$, we see ...


2

Hint: To simply the notation define $\ln_k(x) \equiv \ln\ln\cdots\ln(x)$ where there are $k$ logs on the right hand side. With this notation your integral can be written $$\int^\infty\frac{dx}{x\ln_1(x)\ln_2 (x)\cdots \ln_{k-1}(x)(\ln_k(x))^s}$$ From the definition above it follows that $\ln_{k}(x) = \ln(\ln_{k-1}(x))$ so $$\frac{d\ln_{k}(x)}{dx} = ...


1

You forgot a $(-)$ in an intermediate step but somehow you corrected it in the next step. Your final result is correct. It will not be simplified further (as far as I can see) $(ii)$ To find the coefficients you had $$\frac16 \int ^3 _ 1 12 e^{-in\frac{\pi}{3}x}dx$$ Notice that the integral you want is $n=-3$ (with a factor of $2$ in front of it)


1

Hint: $x^2+y^2\, dV = r^2\,r\,dr\,d\theta\,dz=r^3\,dr\,d\theta\,dz$ in polar coordinates


1

For odd values of n, use the fact that $~\dfrac n{(1+x)(1+x^n)}~=~\dfrac1{(1+x)^2}~+~\dfrac{P_{n-2}(x)}{1+x^n}~,~$ where $P_{n-2}(x)~=~\displaystyle\sum_{k=0}^{n-2}(n-k-1)(-x)^k,~$ in conjunction with the famous identity $~\displaystyle\int_0^\infty\frac{x^{a-1}}{1+x^n}~dx$ $=~\dfrac\pi n\cdot\csc\bigg(a~\dfrac\pi n\bigg),~$ which can be proven using the ...


1

Either you know Riemann-Lebesgue's lemma, and it asserts such a limit is $0$, or you can prove it with in integration by parts: Set $u=\dfrac1{\cos t+4}$, $\mathrm d\mkern1mu v=\cos nt \mathrm d\mkern1mu t$. You obtain: $$\int_{0}^{2\pi} \frac{\cos nt} {\cos t+4}\mathrm d\mkern1mut=\frac{\sin t}{n(\cos y+4)}\Biggr\rvert_0^{2\pi}-\frac1n\int_{0}^{2\pi} ...


1

The easiest way is to apply the long division, but you can make substitution also: let $u=x^2+1\Rightarrow du=2x dx$ Therefore: $\int \frac{x^3+2x}{x^2+1}dx=\frac{1}{2}\int\frac{u+1}{u}du=\frac{u+ln(u)}{2}+C=\frac{x^2+1+ln(x^2+1)}{2}+C=\frac{x^2+ln(x^2+1)}{2}+C$ I don't understand if @mathlove is the first who gives an answer why others give the same ...



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