#Trigonometric substitution
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!da2a
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
it's one specific type of substitution, in essence no different than setting $u=x^2-1$ in an integral like $\int x\sqrt{x^2-1},dx$
elrichardo1337
as for why we do them? if you see some sort of "common-trig-identity-like structure" in an integral a trig sub can simplify it greatly
take for example $\int\frac{dx}{1+x^2}$
elrichardo1337
taking $u=1+x^2$ wouldn't be very helpful here, as since $du=2x,dx,$ we would need an extra factor of $x$ in the numerator that we don't have here
elrichardo1337
but notice the denominator and how it resembles the trig identity $1+\tan^2\theta=\sec^2\theta$
elrichardo1337
so that gives us the idea to set $x=\tan\theta,$ $dx=\sec^2\theta,d\theta$
elrichardo1337
the integral in that case becomes $\int\frac{dx}{1+x^2}=\int\frac{\sec^2\theta,d\theta}{1+\tan^2\theta}=\int\frac{\sec^2\theta,d\theta}{\sec^2\theta}=\int d\theta=\theta+C$
elrichardo1337
but what's $\theta$ in terms of $x?$ take the arctan of our original substitution to get $\theta=\arctan x,$ so indeed this integral is $\arctan x+C$ as expected
elrichardo1337
does that help motivate it a bit better
I see that does make a lot more sense I completely miss understood what we were doing! Now I feel silly since substitution is literally in the name... Thank you so much for the help!
npnp
it's usually taught separately from u sub bc the computations in trig subs tend to be more involved
but fundamentally yeah they're the same thing
.close