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When it’s A square minus f (x) square, f(x) equals to A sin theta

When A square, f(x) square adds up too, f(x) is A tangent theta

When f(x) square minus A square, f(x) equals to A secant theta

Use derivative and identities, substitute all terms to f theta




D arcsin, 1 by square root of 1 minus x square, minus that d arccos

D Arc tangent 1 by 1 plus x square and arc cotangent just minus that

D arc secant is one over x times square root of x square minus 1

These are all the derivatives of the inverse trig functions 




Arc sinhx ln (x plus root x^2 +1), derivative 1 by root x^2 plus one

Arc coshx change second plus to minus, and the same for its derivative 

Arc tanhx half ln (1-x over 1+x), reverse log arc coth

Differentiate them you get 1 by 1 minus x square as your value




Arc sechx just cosh 1 over x, cosechx sinh 1 over x

Former’s derivative minus one over x times root one minus x square

Latter just add absolute sign to the front x and back minus to plus

These are all you need to know for the functions of the hyperbolics




If you need hyperbolic substitution you need to follow the rules below

f x square plus A square you can use f x equals to A sinh theta

Take cosh if f x minus A square, substitute all x so it turns theta

Simplify and then you integrate, you can find the function is the same




If you used trig substitution then you need to draw triangle to get back the x

And if you have definite integral, you need to change limits with inverse trig

Or you can do it directly, but you need to get logs into hyperbolics

This is how you do trig substitutions and you will always do it



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