Integration by Substitution

Integral substitution is one of the integration techniques that is often used in solving integral problems, both indeterminate and integral integrals of course. If substitution integral techniques cannot be done in solving the integration problems, there is a second technique that can be used, namely partial integral techniques. The effectiveness of using this substitution method depends on the availability of a list of known integrals. For this reason, we argue that you must memorize the list of known integrals. Are as follows.

Basic Integral Form

1. $ int k dx = kx + C $
2. $ int x^n dx = frac{x^{n + 1}}{n + 1} + C $
3. $ int frac{1}{x} dx = ln x + C $
4. $ int e^x dx = e^x + C $
5. $ int a^x dx = frac{a^x}{ln a} + C, a neq 1, a> 0 $
6. $ int sin x dx = -cos x + C $
7. $ int cos x dx = sin x + C $
8. $ int sec^2 x dx = tan x + C $
9. $ int csc^2 x dx = -cot u + C $
10. $ int sec x tan x dx = sec x + C $
11. $ int csc x cot x dx = sec x + C $
12. $ int tan x dx = -ln | cos x | + C $
13. $ int cot x dx = ln | sin x | + C $
14. $ int frac{1}{sqrt{a^2-x^2}} dx = sin^{- 1} (frac{x}{a}) + C $
15. $ int frac{1}{a^2 + x^2} dx = frac{1}{a} tan^{- 1} (frac{x}{a}) + C $
16. $ int frac{1}{x sqrt{a^2-x^2}} dx = frac{1}{a} sec^{- 1} (frac{| x |}{a }) + C $
17. $ int sinh x dx = cosh x + C $
18. $ int cosh x dx = sinh x + C $

Substitution in Integral Uncertain
Suppose we face an indeterminate integral in the standard form, just write the answer. If not, look for a substitution that will turn it into a standard form. If the substitution that we try doesn't work, try another one. The ability to do this requires a lot of practice.

Theorem:
Let g be a differentiated function and suppose F is anti-derivative f. Then, if u = g (x),
$ int (g (x)) g'(x) dx = int f (u) du = F (u) + C = F (g (x)) + C $.

Question example:
Look for $ int frac{x}{cos^2 (x^2)} dx $
Settlement:
$ int frac{x}{cos^2 (x^2)} dx = int x sec^2 (x^2) dx $
Suppose that $ u = x^2 $
Then,
$ begin{align} frac{du}{dx} & = 2x \ frac{1}{2} du & = x dx end{align} $
So that,
$ begin{align} int frac{x}{cos^2 (x^2)} dx & = int x sec^2 (x^2) dx \ & = int sec^2 (x^2) x dx \ & = int sec^2 (u) frac{1}{2} du \ & = frac{1}{2} int sec^2 (u) du \ & = frac{1}{2} tan u + C \ & = frac{1}{2} tan (x^2) + C end{align} $
Substitution in Integral Naturally
This is like substitution in indeterminate integrals, but we must remember to make appropriate changes in integration limits.
Question example:
Calculate $ int_{2}^5 t sqrt{t^2-4} dt $
Settlement:
Suppose that $ u = t^2-4 $
Then,
$ begin{align} frac{du}{dt} & = 2t \ frac{1}{2} du & = t dt end{align} $
$ begin{align} t = 2 rightarrow u & = (2)^2-4 \ u & = 0 end{align} $
$ begin{align} t = 5 rightarrow u & = (5)^2-4 \ u & = 21 end{align} $
So that,
$ begin{align} int_{2}^5 t sqrt{t^2-4} dt & = int_{2}^5 sqrt{t^2-4} t dt \ & = int_{0}^{21} sqrt{u} frac{1}{2} du \ & = frac{1}{2} int_{0}^{21} u^{frac{1}{2}} du \ & = frac{1}{2} (frac{2}{3} u^{frac{3}{2}})]_{0}^{ 21} \ & = frac{1}{3} u sqrt{u}]_{0}^{21} \ & = frac{1}{3} (21) sqrt{21} - frac{1}{3} (0) sqrt{0} \ & = 7 sqrt{21} end{align} $
The substitution discussed is the first version of substitution, there is a second version substitution. To find out the example of the second version of the substitution method.