2.6.62 inverse integrals with substitution

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karush
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ok this is from my overleaf doc
so too many custorm macros to just paste in code
but I think its ok,,, not sure about all details.
appreciate comments...

I got ? somewhat on b and x and u being used in the right places
 
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  • Handwritten-style typeset solutions to two integrals: first, the integral of dx over the square root of 1 minus x squared times arcsin x, solved by substituting u = arcsin x to give ln|arcsin(x)| + C; second, the integral from 0 to sqrt(3)/4 of dx over 1 + 16x squared, solved using arctan with a = 1, u = 4x, giving pi/12.
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Last edited:
karush said:
ok this is from my overleaf doc
so too many custorm macros to just paste in code
but I think its ok,,, not sure about all details.
appreciate comments...

I got ? somewhat on b and x and u being used in the right places
Ummm...

What exactly is your question?

-Dan
 
OK this one was kinda obvious but usually that is where I make the errors especially where there is substitution.
I post another one here soon.:cool:
 
karush said:
That all looks correct to me, except that at one point in the solution of 6.6.62 you have written $du=dx$. That should be $\dfrac{du}4 = dx$ (which is what you have correctly used in the following line).
 
Attachments
  • Handwritten calculus solutions to integrals 6.6.65 and 6.6.62 with u-substitution steps, ending in ln|arcsin x| + C and pi/12.
    253a.PNG
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