Now all these integrals I post aren't in vain! The reason I do this is because I am conducting a study/project for school that in the long term will hopefully produce a booklet containing the most down to earth intuitive integration techniques known to highschooldom!

So I am going to do a couple of integrals..and if you see a more intutive way of doing them I would really appreciate it if you would post your way! Thanks so much...all in this work in the name of math

We will start easy...how intuitive do you think the method I always preach on posts of getting integrals into $\displaystyle \int{f(g(x))\cdot{g'(x)}dx}$ by constant manipulation is?

For example if I wrote something to the effect of

$\displaystyle \int{t\sqrt{t^2+1}dt}=\frac{1}{2}\int{2t\sqrt{t^2+ 1}}dt$

$\displaystyle \frac{1}{3}(t^2+1)^{\frac{3}{2}}+C$

and now I will just do a couple without a small oration before hand

$\displaystyle \int\frac{1}{1+\tan(x)}dx$

Let $\displaystyle u=\tan(x)\Rightarrow{arctan(u)=x}$

Then $\displaystyle dx=\frac{1}{1+u^2}du$

So we have $\displaystyle \int\frac{1}{(1+u)(1+u^2)}du$

Using PFD we see that $\displaystyle \frac{1}{(1+u)(1+u^2)}=\frac{1}{2(u+1)}+\frac{1}{2 (u^2+1)}-\frac{u}{2u^2+2}$

Integrating we get $\displaystyle \frac{1}{2}\ln|u+1|+\frac{1}{2}arctan(u)-\frac{1}{4}\ln|x^2+1|$

Back subbing we get

$\displaystyle \int\frac{1}{1+\tan(x)}dx=\frac{1}{2}\ln|1+\tan(x) |+\frac{x}{2}+\frac{1}{2}\ln|\cos(x)|+C$

The next one is

$\displaystyle \int\sqrt{1-e^{x}}dx$

Let $\displaystyle u=\sqrt{1-e^{x}}\Rightarrow{\ln(1-u^2)=x}$

THen $\displaystyle dx=\frac{-2u}{1-u^2}du$

So we have $\displaystyle \int\frac{-2u^2}{1-u^2}du=\int\frac{-2u^2+2}{1-u^2}du-2\int\frac{1}{1-u^2}$$\displaystyle =2\int\frac{1-u^2}{1-u^2}du-2\int\frac{1}{1-u^2}$

by PFD we have $\displaystyle \int{2}du+\int\frac{1}{u-1}du-\int\frac{1}{u+1}du$

Integrating we get $\displaystyle 2u+\ln|u-1|-\ln|u+1|+C$

Making our back sub we get $\displaystyle 2\sqrt{1-e^{x}}+\ln|\sqrt{1-e^{x}}-1|-\ln|\sqrt{1-e^{x}}+1|+C$

The last one is

$\displaystyle \int\frac{sin(4x)}{16+\cos^4(2x)}dx$

Letting $\displaystyle u=\cos^2(2x)$

THen $\displaystyle du=2\cos(2x)\sin(2x)=\sin(4x)$

Subbing we have

$\displaystyle \int\frac{du}{16+u^2}$

Which then yields $\displaystyle \frac{1}{4}arctan\bigg(\frac{u}{4}\bigg)+C$

Back subbing we get

$\displaystyle \int\frac{\sin(4x)}{16+\cos^4(2x)}dx=\frac{1}{4}ar ctan\bigg(\frac{\cos^2(2x)}{4}\bigg)+C$

So if anyone sees an easier way to do these please do tell! I need as much imput as possible for this booklet