Inverse Trig (1 Viewer)

[azureus88]

How would you set out the integration of:

int [6/(49+25x^2)]dx

Are we meant to use the standard forms or use substitutions?

 

[lyounamu]

How would you set out the integration of:

int [6/(49+25x^2)]dx

Are we meant to use the standard forms or use substitutions?

int (6/(25(49/25+x^2)))dx
= 6/25 int (1/(49/25+x^2) dx
= 6/25 x 5/7 int ((7/5)/(49/25 + x^2))dx
= 6/35 tan-1(5/7 . x) + c

 

[Pupzrollin]

= 6/35 tan-1(5/7x) + c[ Hey Namu i think your last line's a wee bit wrong as it should be *6/35 tan-(5x/7) +c* as the formula is tan-1(x/a) and a is *7/5*.]

 

[lyounamu]

= 6/35 tan-1(5/7x) + c[ Hey Namu i think your last line's a wee bit wrong as it should be *6/35 tan-(5x/7) +c* as the formula is tan-1(x/a) and a is *7/5*.]

Well I was meant 5/7 . x by 5/7x that's why there was no bracket.

haha, anyway.

edit:fixed

 

[Timothy.Siu]

How would you set out the integration of:

int [6/(49+25x^2)]dx

Are we meant to use the standard forms or use substitutions?

u cud use substitution but its just slower

 

[azureus88]

thanks ppl! but how did u do this bit in ur head?

6/25 int (1/(49/25+x^2) dx
= 6/25 x 5/7 int ((7/5)/(49/25 + x^2))dx

 

[Pupzrollin]

thanks ppl! but how did u do this bit in ur head?

6/25 int (1/(49/25+x^2) dx
= 6/25 x 5/7 int ((7/5)/(49/25 + x^2))dx

Well...Firstly, you have to take the co-efficient of 25 (denominator) out from 25x^2 and 6 from the top as the numerator to make it in the form [ a/ a^2 +x^2] because ultimately Ur result will be (1/a) tan-1(x/a) +c. Your "a" becomes 7/5 from the (7/5)^2 from the denominator and you must make the numerator (7/5) also. Thus, to make the numerator *7/5* you must multiply the coefficient of the integral by *5/7* to make everything equal. And by using the formula u'll get the desired answer.