Integration by Substitution (1 Viewer)

[Crosswinds]

I'm stuck on this integration question:

Find ∫ dx / (25 + x²)^3/2 by letting x = 5tan@

I keep getting 1/25 sin(tan^-1(x/5)) + C but the answers give 1/25 x/√(25 + x²) + C.

Are these answers equivalent (if so could someone please show me why) or am I making a stupid mistake?

Please and thankyou!

 

[tommykins]

I'm stuck on this integration question:

Find ∫ dx / (25 + x²)^3/2 by letting x = 5tan@

I keep getting 1/25 sin(tan^-1(x/5)) + C but the answers give 1/25 x/√(25 + x²) + C.

Are these answers equivalent (if so could someone please show me why) or am I making a stupid mistake?

Please and thankyou!

yeah they are equivalent

draw a triangle and ull see it

 

[dog on heat]

man i could go for some of this riht now

 

[madsam]

I'm stuck on this integration question:

Find ∫ dx / (25 + x²)^3/2 by letting x = 5tan@

I keep getting 1/25 sin(tan^-1(x/5)) + C but the answers give 1/25 x/√(25 + x²) + C.

Are these answers equivalent (if so could someone please show me why) or am I making a stupid mistake?

Please and thankyou!

They are equivalent, however it isn't in the most simplified form

after you've done the substitution and everything, you end up with

1/25 . sin@
Where x = 5tan@

So what you have done, is made that:
tan^-1 (x/5) = @ and subbed that in

HOWEVER, the alternative (and most correct) answer, which is what they used is:
Using your x/5 = tan@, you draw a tight angle triangle, label two of the sides x and 5 (not the hypotenuse) and then using tan = opp/adj label your @ angle

Now you need to find sin@, and you need the hypotenuse for this, and simply use pythagoras and you get x/(sqrt(x^2 + 25)

you just sub that in

1/25 . x/sqrt(x^2 + 25)

 

[Crosswinds]

Ohhhh yeaah. Okay thanks for your help!