Question

Inverse trig function?

Prove by means of differentiation that if a > 0

Answer 1

`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

Answer 2

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`

Answer 3

`"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`