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

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

where:

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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