`int cosx/sqrt(sin^2x+1) dx` Use integration tables to find the indefinite integral.

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Recall that indefinite integral follows: `int f(x) dx = F(x)+C`


`f(x)` as the integrand function

`F(x) ` as the antiderivative of `f(x)`

`C` as constant of integration.

To evaluate given integral problem: `int cos(x)/sqrt(sin^2(x)+1)dx` or `int (cos(x)dx)/sqrt(sin^2(x)+1)` , we may apply u-substitution by letting:

`u = sin(x)` then `du =cos(x) dx` .

Plug-in the values ,  the integral becomes:

`int (cos(x)dx)/sqrt(sin^2(x)+1)=int (du)/sqrt(u^2+1)` or `int (du)/sqrt(u^2+1^2)`

The integral resembles one of the formulas from the integration table for rational function with roots. We follow:

`int (dx)/sqrt(x^2+a^2) = ln|x+sqrt(x^2+a^2)|+C`

By comparing `x^2+a^2` with `u^2+1^2` , we determine the corresponding values as: x=u and a=1.

Applying the values on the integral formula for rational function with roots, we get:

`int (du)/sqrt(u^2+1^2)=ln|u+sqrt(u^2+1^2)| +C`

                                `=ln|u+sqrt(u^2+1)| +C`

Plug-in `u = sin(x)` on  `ln|u+sqrt(u^2+1)| +C` , we get the indefinite integral as:

`int cos(x)/sqrt(sin^2(x)+1)dx=ln|sin(x)+sqrt(sin^2(x)+1)| +C`


 Aside from this, we can also consider the another formula from integration table:

`int 1/sqrt(u^2+1)du = arcsinh(u) +C`

Plug-in `u = sin(x) ` on `arcsinh(u) +C` , we get another form of indefinite integral as:

`int cos(x)/sqrt(sin^2(x)+1)dx=arcsinh(sin(x)) +C`


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