How calculus teachers create integrals for your test!

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How calculus teachers create integrals for your test!

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This is how the last boss in the game transform into its final phase.

nattoro

Dr : there is a bonus question make sure u do it
The bonus question:

faisal_tarshon

That d/dx actually makes this slightly easier since all of the ln’s become fractions

alberteinstein

Can't believe it's solvable...
sin²x + cos²x + tan²x = 1 + tan²x = sec²x

Bottom differentiation is equal to -2cotx at last (too much to write)

So, the whole thing:
=§(1+ln(sec²x)^π) / -2cotx) dx
=(-1/2) §tan x (1+ln(sec²x))^π dx

Now let u = 1+ln(sec²x)
du/dx = 2tanx
dx= du/2tanx

So now change everything to the "u" world:
=(-1/2) § u^π (tan x) / 2tan x du
tan x cancels, take 1/2 out:
=-1/4 §u^πdu
Integrate and put back what u is:
=-(1+ln(sec²x))^(π+1) / 4π+4 +C

Am I right?...

yunamiyue

All in one question. This question solve using all of technique in integrations.

aldrinponce

Imagine if the dramatic bass sounds (the one with the rock tilting his head), playing every time he ln or square the number or add a fraction.

zacharymcguirk

The d/dx makes the whole denominator of ln into fractions. The sum of squares of sine and cosine is 1, so the numerator becomes (2 + tan^2 x)^π which is just (1 + sec^2 x)^π. So all that extra brainstorming all but made it easier to solve 😂

mannatchaudhary