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0

We have that $$\int_{\pi/4}^{\pi/2} \frac{x \cos(x)-\sin(x)}{x} dx=\int_{\pi/4}^{\pi/2}\bigg( \cos(x)-\frac{\sin(x)}{x}\bigg) dx$$ where $$\int_0^x\frac{\sin t}{t}\,dt={\text{Si}}(x) $$ is the Sine integral. Hence \begin{align}\int_{\pi/4}^{\pi/2}\bigg( \cos(x)-\frac{\sin(x)}{x}\bigg) dx&=\int_{\pi/4}^{\pi/2}\cos(x)dx ~-~\int_{\pi/4}^{\pi/2}\frac{\...


1

Too long for a comment. You could have done it faster using $$\sqrt{1 + \tan(x)}=t \implies x=-\tan ^{-1}\left(1-t^2\right)\implies dx=\frac{2 t}{1+\left(1-t^2\right)^2} \,dt$$ making $$I=\int \sqrt{1 + \tan(x)}\,dx=\int \frac{2 t^2}{1+\left(1-t^2\right)^2} \,dt$$ Using now $$1+\left(1-t^2\right)^2=(t^2-(1-i))(t^2-(1+i))$$ and using partial fraction ...


2

I know why you don't want to go over that again. :) Why don't you instead make the substitution $$1+\tan x=u^2,$$ to obtain $$2\int{\frac{u^2}{1+(u^2-1)^2}\mathrm d u},$$ which may be easily done by parts, as follows, for example, $$u\int{\frac{2u}{1+(u^2-1)^2}\mathrm d u}-\int\cdots,$$ where the integral in the first summand is easy to do by using the ...


2

When you substitute $U=\sqrt{x^3-6}$, you made a mistake. It should be $$\int{\dfrac{2U}{U\sin^2{U}}dU}=\int{2\csc^2{U}dU}$$, so the answer is $$-2\cot^2{\sqrt{x^3-6}}+C$$


3

$$u=\sqrt{x^3-6},du=\dfrac{3x^2}{2\sqrt{x^3-6}}dx$$ $$2\int\csc^2u=-2\cot u+K$$


1

I suggest that you write $$ \frac{1}{(u^2+a^2)^{m-1}} = u\cdot\frac{u}{(u^2+a^2)^m}+\frac{a^2}{(u^2+a^2)^m}, $$ and integrate by parts in the first term in the right-hand side, and then rearrange your terms.


1

Using standard techniques to calculate infinite series via residue calculus one obtains that, if $$ \pi \cot (\pi z)=\frac{a_{-1}}{z}+a_1z+a_3z^3+\cdots+a_{2k-1}z^{2k-1}+\cdots $$ then $$ \sum_{n=1}^\infty \frac{1}{n^{2k}}=-2a_{2k-1}. $$ Hence $$ \sum_{n=1}^\infty \sin^2\left(\frac{\pi}{n}\right)=\frac{1}{2}\sum_{n=1}^\infty \left(1-\cos\Big(\frac{2\pi}{n}\...


5

$\sum_{n=1}^\infty \sin^2(\frac{\pi}{n}) = \sum_{n=1}^\infty \frac{1-\cos(\frac{2\pi}{n})}{2} = \frac{1}{2}\sum_{n=1}^\infty \left[\frac{4\pi^2}{2n^2}-\frac{2^4\pi^4}{4!n^4}+\frac{2^6\pi^6}{6!n^6}+\dots\right] = \frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\frac{2^{2n}\pi^{2n}}{(2n)!}\zeta(2n) = \frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\frac{2^{2n}\pi^{2n}}{(2n)!}(-...


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For a) one possible solution is given by $$\arctan\frac{x}{\sqrt{a^2-x^2}}+C$$ For b) it is as you stated $$\frac{\arctan\left(\frac{x}{a}\right)}{a}+C$$


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Use the property that $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx$$ Let $$I=\int_{0}^{\pi/2} \frac{\cos^{n}(x)}{\sin^{n}(x)+\cos^{n}(x)} \ dx$$ Then by our property we have $$I=\int_{0}^{\pi/2} \frac{\sin^{n}(x)}{\sin^{n}(x)+\cos^{n}(x)}\ dx$$ Add both


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