# Evaluate the integral `int_0^(1/sqrt(2)) arccos x /sqrt(1-x^2) dx`

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### 3 Answers

To evaluate this integral, notice that a component of the integrand is the term

`1/sqrt(1-x^2)` . When a term of this type appears in the integrand it is usually a good idea to make a sine or cosine trigonometric substitution.

In the numerator of the integrand is an arccos(x) term, where the arccos function is the inverse cosine function. This suggests that a substitution of the form

`x = cos theta`

would be useful, as this arccos(x) term would then become a constant.

So, make the substitution

`x = cos theta` so that `theta = arccos(x)` and `dx = -sin theta d theta`

The limits on the (definite) integral need to be transformed to the `theta` scale. If we have `x in [0,1/sqrt(2)]` this implies that `theta in [pi/4, pi/2]`

since `arccos(0) = pi/2` and `arccos (1/sqrt(2)) = pi/4` .

Therefore, putting this all together we can rewrite the integral as

`I = int_0^(1/sqrt(2)) (arccos(x))/sqrt(1-x^2) dx` `= int_(pi/4)^(pi/2) (-theta sin theta)/sqrt(1-cos^2(theta)) d theta`

**Using the trig identity `sin theta = sqrt(1- cos^2(theta))` we then have that**

**`I = int_(pi/4)^(pi/2) - theta d theta = - theta^2/2|_(pi/4)^(pi/2) = -pi^2/32 + pi^2/8 = (3pi^2)/32` **

**Sources:**

Another way to do this is to make a u substitution.

`int_0^(1/sqrt(2)) arccos(x)/sqrt(1-x^2) dx`

let `u = arccos(x)`

`du = -1/sqrt(1-x^2) dx`

We can substitute that back into our integral.

`int -u du`

`-u^2/2`

`-(arccos^2(x))/2` from 0 to 1/sqrt(2)

And plug that in, and we get `(3pi^2)/32`

the solution is 1/2 cosh^(-1) x^2 with the limits 0 and 1/(2)^0.5

solving completely, we get the answer as 0.5235 - 0.785i.

good luck.