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 KD3203 二星級會員 帖子253 積分377 註冊時間2017-1-1 Trigonometric and hyperbolic formulas and substitutions song (原曲：屯馬開通真的很興奮) #1 發表於 2021-10-3 09:48 PM  只看該作者 繁 簡 [按此打開] [隱藏] 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 .postmorevideochannel { margin-top: 10px; display: block; width: 160px; height: 30px; background: url('/images/video_channel_more.png?v=20160829b') no-repeat; font-size: 0; } .postmorevideochannel:hover, .postmorevideochannel:focus { background-position: 0 -30px; } @media (-webkit-min-device-pixel-ratio: 2), (min-resolution: 192dpi) { .postmorevideochannel { background: url('/images/video_channel_more_2x.png?v=20160829b') no-repeat; background-size: 100% auto; } }