Inverse trig function?

Prove by means of differentiation that if a > 0
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3 Answers
Aug 3, 2017

int \ 1/(a^2+u^2) \ du = 1/a arctan(u/a) + C

Explanation:

We want to show:

I = int \ 1/(a^2+u^2) \ du = 1/a arctan(u/a) + C

Method 1 - Differentiation

Let us denote the function by y. as follows::

y =1/a arctan (u/a)
:. ay =arctan (u/a)
:. tan(ay) =u/a

Implicitly differentiating we get:

asec^2(ay)dy/(du) = 1/a
:. sec^2(ay)dy/(du) = 1/a^2

Using the trig Identity tan^2A+1=sec^2A we have

:. (tan^2(ay)+1)dy/(du) = 1/a^2
:. ((u/a)^2+1)dy/(du) = 1/a^2
:. dy/(du) = 1/( a^2 (1+(u/a)^2) )
:. dy/(du) = 1/( a^2 (1+u^2/a^2 )
:. dy/(du) = 1/(a^2+u^2)

Hence we have:

d/du(1/a arctan(u/a) ) = 1/(a^2+u^2)
=> int \ 1/(a^2+u^2) \ du = 1/a arctan(u/a) + C \ \ \ QED

Method 2 - Integration

Although asked to differentiate to form the solution, we can also readily integrate the function:

Let us perform a substitution:

u=a tan theta => (du)/(d theta) = asec^2 theta; \ \ theta = arctan(u/a)

Then provided a != 0 substituting into the integral gives:

I = int \ 1/(a^2+a^2tan^2theta) \ (asec^2 theta) \ d theta
\ \ = 1/a \ int \ (sec^2 theta)/(1+tan^2theta) \ d theta
\ \ = 1/a \ int \ (sec^2 theta)/(sec^2theta) \ d theta
\ \ = 1/a \ int \ d theta
\ \ = 1/a theta + C

And restoring the substitution, we get:

I = 1/a arctan(u/a) + C \ \ \ \ QED

Aug 3, 2017

See below.

Explanation:

d/dx(arctan(x)) = 1/(1+x^2)

Now use the chain rule. Note that a is a constant.

d/(du)(1/aarctan(u/a)) = 1/a(1/(1+(u/a)^2) * d/(du)(u/a))

= 1/a(1/(1+(u/a)^2) * (1/a))

= 1/(a^2+u^2)

Therefore,

int 1/(a^2+u^2) =1/aarctan(u/a) +C

Note

We don't need the restriction a > 0. We only need a != 0

Aug 3, 2017

"see explanation"

Explanation:

y=1/atan^-1(u/a)+c

rArrdy/(du)=1/axx1/(1+(u/a)^2)xxd/(du)(u/a)

color(white)(rArrdy/du)=1/a^2xx1/(1+(u^2/a^2)

color(white)(rArrdy/du)=1/(a^2+u^2

rArrint(du)/(a^2+u^2)=1/atan^-1(u/a)+ctoa>0