[Not a Question][Link included] Answers to my integration thread

Link:https://web2.0calc.com/questions/loooong-time-no-see

Seeing that no one did my integrals, I will reveal answers,

1)

\(\int \dfrac{\tan^3 \theta \sec \theta d\theta}{\sqrt{\tan^2 \theta + 2}}\\ =\int \dfrac{(\sec^2 \theta - 1)(\sec\theta \tan \theta d\theta)}{\sqrt{\sec^2\theta + 1}}\boxed{u=\sec \theta\\ du=\sec\theta\tan\theta d\theta}\\ =\int \dfrac{u^2 - 1}{\sqrt{u^2+1}}du\boxed{u=\tan\phi\\du=\sec^2\phi d\phi}\\ =\int \dfrac{\tan^2\phi - 1}{\sec\phi}d\phi\\ =\int\sec \phi d\phi -2\int \cos \phi d\phi\\ =\ln|\sec\phi + \tan \phi|-2\sin \phi + C\\ =\ln|\sqrt{u^2 + 1}+u|-\dfrac{2u}{\sqrt{u^2+1}}+C\\ =\ln|\sqrt{\tan^2\theta+2}+\sec \theta|-\dfrac{2\sec \theta}{\sqrt{\tan^2\theta + 2}}+C \)

2)

\(\int \dfrac{\sec2\theta(\cos \theta - \sin \theta)^2}{\sqrt{1+\sin 2\theta}}d\theta\\ =\int \dfrac{(\cos \theta - \sin \theta)^2}{(\cos^2\theta - \sin^2\theta)(\sin \theta + \cos \theta)}d\theta\\ =\int \dfrac{(\cos \theta - \sin \theta)(\cos \theta - \sin \theta)}{(\cos \theta - \sin \theta)(\sin \theta + \cos \theta)(\sin \theta + \cos \theta)}d\theta\\ =\int\dfrac{(\cos \theta - \sin \theta)}{(\sin \theta + \cos \theta)^2}d\theta \boxed{u=\sin \theta + \cos \theta\\ du=(\cos \theta - \sin \theta)d\theta}\\ =\int u^{-2}du\\ =-u^{-1} + C\\ =-\dfrac{1}{\sin \theta + \cos \theta}+C\)

3)

\(\int \sqrt{\cot x}-\sqrt{\tan x}\;dx\\ =\int \dfrac{\sqrt{\cos x}}{\sqrt{\sin x}}-\dfrac{\sqrt{\sin x}}{\sqrt{\cos x}}dx\\ =\int \dfrac{\cos x - \sin x}{\sqrt{\sin x \cos x}}dx\\ =\sqrt2\int \dfrac{\cos x - \sin x}{\sqrt{\sin 2x}}dx\\ =\sqrt2\int \dfrac{\cos x - \sin x}{\sqrt{(\sin x + \cos x)^2 -1}}dx\;\boxed{u=\sin x + \cos x\\du=(\cos x - \sin x)dx}\\ =\sqrt2\int \dfrac{du}{\sqrt{u^2-1}}\\ =\sqrt2\cosh^{-1}u + C\\ =\sqrt2\cosh^{-1}(\sin x + \cos x)+ C\)

4) This is a special one made by me, and is too hard for anyone.

You need to know the derivatve and second derivative of x^x to do this integral.

\(\int x^{x+1}(\ln x + 1)^2 - x^x \;dx\\ = \int x(x^x(\ln x + 1)^2 - x^{x-1})dx\boxed{u=x\quad dv=(x^x(\ln x + 1)^2-x^{x-1})dx\\du=dx\quad v=x^x(\ln x + 1)}\\ =x^{x+1}(\ln x + 1)-\int x^x(\ln x + 1)dx\\ = x^{x+1}(\ln x + 1)-x^x + C\)

5) From YT channel "blackpenredpen". Thank him for teaching us how to do 5) and 6).

\(\int \sqrt{\tanh x}dx\boxed{u=\sqrt{\tanh x}\quad u^2 = \tanh x\quad 2udu=\text{sech}^2xdx\\dx=\dfrac{2u}{1-u^4}du}\\ =\int \dfrac{2u^2}{1-u^4}du\\ =\int \dfrac{1}{1-u^2}-\dfrac{1}{1+u^2}du\\ =\tanh^{-1}u -\tan^{-1}u + C\\ =\tanh^{-1}(\sqrt{\tanh x}) - \tan^{-1}(\sqrt{\tanh x})+ C\)

6) Also from "blackpenredpen".

\(\int \sqrt[3]{\tan x}dx\boxed{u=\sqrt[3]{\tan x}\quad 3u^2du = \sec^2 x dx}\\ =3\int \dfrac{u^3}{1+u^6}du\boxed{t = u^2}\\ =\dfrac{3}{2}\int \dfrac{t}{1+t^3}dt\\ =-\dfrac{1}{2}\int \dfrac{dt}{1+t}+\dfrac{1}{2}\int \dfrac{t+1}{t^2-t+1}\\ \boxed{\color{red} \text{I don't show steps for this for space}}\\ =-\dfrac{1}{2}\ln|1+\tan^{2/3}x|+\dfrac{1}{4}|\tan^{4/3}x-\tan^{2/3}x+1|\\\qquad +\dfrac{\sqrt3}{2}\arctan(\dfrac{2\tan^{2/3}x-1}{\sqrt3})+C \)

7)

\(\int \dfrac{1+\sin \theta + \cos \theta}{1-\sin \theta + \cos \theta}d\theta\boxed{u=\tan\dfrac{\theta}{2}\quad d\theta = \dfrac{2}{1+u^2}du\\\sin\theta = \dfrac{2u}{1+u^2}\\\cos \theta = \dfrac{1-u^2}{1+u^2}} \\ =\int \dfrac{1+u^2+2u+1-u^2}{1+u^2-2u+1-u^2}\cdot \dfrac{2du}{1+u^2}\\ =2\int \dfrac{u+1}{1-u}\cdot \dfrac{du}{1+u^2}\\ =\int \dfrac{2u}{u^2+1}du-2\int\dfrac{1}{u-1}\\ =\ln|u^2+1|-2\ln|u-1|+C\\ =2\ln|\sec\dfrac{\theta}{2}|-2\ln|\tan\dfrac{\theta}{2}-1|+C\)

8) Actually this is the easiest among all 9 integral I provided.

\(\int \dfrac{x^3}{\sqrt{x^8+1}}dx\boxed{u = x^4\quad du = 4x^3dx}\\ =\dfrac{1}{4}\int\dfrac{du}{\sqrt{u^2+1}} \\ =\dfrac{1}{4}\sinh^{-1}u+C\\ =\dfrac{1}{4}\sinh^{-1}(x^4)+C\)

9) Uses the result of 3) and integral of sqrt tan x.

Know that: \(\int \sqrt{\tan x}dx = \dfrac{1}{\sqrt2}\arctan(\dfrac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt2})\\ \qquad\qquad\quad-\dfrac{1}{\sqrt2}\tanh^{-1}(\dfrac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt2})+C\)

So:

\(\int \sqrt{\cot x}dx\\ =\int (\sqrt{\cot x} - \sqrt{\tan x}) dx +\int \sqrt{\tan x}dx\\ =\sqrt2 \cosh^{-1}(\sin x + \cos x)-\dfrac{1}{\sqrt2}\tanh^{-1}(\dfrac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt2})\\ \qquad +\dfrac{1}{\sqrt2}\arctan(\dfrac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt 2}) + C\)