Question

How do we know if our substitution will work in integration

Answer 1

we wouldn't know.
we just have to try and solve it.
if it doesn't work, we have to try new substitution.
no specific method to ensure correct substitution.

Answer 2

Okay, then basicaly trial and error

Answer 3

there are some standard substitutions for standard expressions.
if you know them, most of your substitution problems will get covered.

Answer 4

let me pull out that table i made ...

Answer 5

Fine!

Answer 6

\(\begin{array}{|c|c|}\hline \text{Expression in Integral} &Substitution
\\ \hline \sqrt{a^2-x^2}&x=a\sin \theta \quad or \quad x=a\cos\theta
\\ \hline \sqrt{x^2-a^2}&x=a\sec \theta \quad or \quad x=a\csc\theta
\\ \hline x^2+a^2 &x=a\tan \theta \quad or \quad x=a\cot\theta
\\ \hline \sqrt{\frac{a-x}{a+x}}& x=a\cos2\theta
\\ \hline \sqrt{\frac{a-x}{x}}or\sqrt{\frac{x}{a-x}} & x=a\sin^2\theta
\\ \hline \sqrt{\frac{a+x}{x}}or\sqrt{\frac{x}{a+x}} & x=a\tan^2\theta
\\ \hline \sqrt{2ax+x^2} & x=2a\tan^2\theta
\\ \hline \sqrt{2ax-x^2} & x=2a\sin^2\theta
\\ \hline \sqrt{\frac{a^2-x^2}{a^2+x^2}} & x^2=a^2\cos2\theta


\\ \hline \end{array}\)

Answer 7

There are so some extremely wacky substitutions you might come across. There could also be problems where substitution fails to help.

Answer 8

\(\begin{array}{|c|c|}\hline \text{Expression in Integral} &Substitution
\\ \hline \ln|f(x)| & u=ln|f(x)|
\\ \hline \ln|f(x)|\pm \ln|g(x)| & u=ln|f(x)| )|\pm \ln|g(x)|
\\ \hline f(x)^nf’(x) & u=f(x)
\\ \hline e^{f(x)} \quad or \quad a^{f(x)} & u=f(x)

\\ \hline \sqrt{ax+b} \\ \frac{cx+d}{\sqrt{ax+b} }\\(cx+d) \sqrt{ax+b} & u= \sqrt{ax+b}
\\ \hline \frac{\sin \:x+\cos \:x}{a+b\sin\:2x} & u=\int Numerator
\\ \hline P(x)(ax+b)^n \\ \text{P(x)is any polynomial in x} & u=ax+b
\\ \hline \frac{1}{x^{1/m}+x^(1/n)} & x=t^k,k=LCM(m,n)



\\ \hline \end{array}\)

Answer 9

oh and the famous substitution :
\(\\~ \\~ \\~ \\~

\\ \text{1. To integrate} \huge \frac{1}{a\sin\:x+b\cos\:x+c}\\ \text{put, t=tan(x/2),then} \large \sin\:x =\frac{2t}{1+t^2} \quad \cos\:x=\frac{1-t^2}{1+t^2} \quad dx=\frac{2}{1+t^2}
\\\)

Answer 10

excellent , so i suppose to memorise, however i feel the intuition to what to substitute comes with experience as my teacher said

Answer 11

no, you solve a couple of problems for each of them and you won't feel the necessity to memorize.

Answer 12

Look for a part of the function which you can integrate..

Answer 13

Okay, I am still a beginner in the calculus world.
Thanks a million

Answer 14

You should try this one:
Isn't really hard but it's neat stuff
\[\int\limits_{}^{}\frac{1}{x^7+x} dx\]

Answer 15

integrate that part of it, and see if the anti-derivative appears in the integral as well.

Answer 16

And yes it can be done by a substitution

Answer 17

Another fun one and maybe not so straight forward one:
\[\int\limits_{}^{}x^3 \sqrt{1-x^2} dx\]

Answer 18

there are substitutions that are pretty straight forward
these two I recommend you do because they do involve a little creativity