Where do I find the Integration of a particular function Formula?

Below are a few important tips to remember the integrals of some particular functions:
1. Know the derivatives for each integral.
2. Make yourself familiar with the standard formulas first.
3. Practice daily for better memory.
4. Group similar formulas

${\displaystyle \int \frac{\left(sinx-cosx\right)si{n}^{2}x}{tanx\left(si{n}^{3}x+co{s}^{3}x\right)}dx}$

${\displaystyle \int \frac{\left(sinx-cosx\right)sinxcosx}{si{n}^{3}x+co{s}^{3}x}dx}$ , put sin³x + cos³x = t(3 sin²x×cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3}{\displaystyle \int \frac{dt}{t}}$

$=\frac{lnt}{3}+c$

$=\frac{ln\left|si{n}^{3}x+co{s}^{3}x\right|}{3}+c$

$\underset{h\to 0}{lim}\frac{{\displaystyle {\int }_{\left(\frac{\pi }{2}-h\right)}^{{\left(\frac{\pi }{2}\right)}^3}cos\left(t^{1/3}\right)dt}}{h^2}$

$=\underset{h\to 0}{lim}\frac{0+3{\left(\frac{\pi }{2}-h\right)}^{2}cos\left(\frac{\pi }{2}-h\right)}{2h}$

$=\underset{h\to 0}{lim}\frac{3{\left(\frac{\pi }{2}-h\right)}^{2}sinh}{2h}$

$=\frac{3{\pi }^2}{8}$

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}}$ dx

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left\{\left(\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}+\frac{8\sqrt[]{2}cosx}{\left(1+e^{-sinx}\right)\left(1+si{n}^{4}x\right)}\right)\right\}dx}$            

$=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{cosx}{1+si{n}^{4}x}dx}$            

Let sin x = t

$I=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^4}}$            

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}}dt}$       

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)dt}{{\left(t-\frac{1}{t}\right)}^2+2}}-4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1-\frac{1}{t^2}\right)dt}{{\left(t+\frac{1}{t}\right)}^2-2}}$            

$=4\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}{\left(ta{n}^{-1}\frac{t-\frac{1}{t}}{\sqrt[]{2}}\right)}_0^1-4\sqrt[]{2}\cdot \frac{1}{2\sqrt[]{2}}{\left[log\left|\frac{t+\frac{1}{t}-\sqrt[]{2}}{t+\frac{1}{t}+\sqrt[]{2}}\right|\right]}_0^1$         

$=2\pi -2log\left|\frac{2-\sqrt[]{2}}{2+\sqrt[]{2}}\right|$       &nbs

$I={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\frac{xdx}{si{n}^4\left(2x\right)+co{s}^4\left(2x\right)}}$

           Let 2x = t then   $dx=\frac{1}{2}dt$

$I={\displaystyle \underset{}{\overset{}{\int }}\frac{\frac{t}{2}\cdot \frac{1}{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{tdt}{si{n}^{4}t+co{s}^{4}t}dt}$            

$I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{\pi }{2}-t\right)dt}{si{n}^{4}t+co{s}^{4}t}dt}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{1}{4}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\pi }{2}dt}{si{n}^{4}t+co{s}^{4}t}}$

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{si{n}^{4}tdt}{ta{n}^{4}t+1}}$            

Let tan t = y then

$2I=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+y^2\right)dy}{1+y^4}}$             

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2}dy}$

$=\frac{\pi }{8}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\left(1+\frac{1}{y^2}\right)dy}{2+{\left(y-\frac{1}{y}\right)}^2}}$             

Let