Question

∫{cosxdx}{\sqrt{1+sin^{2}x } }" alt="\frac{cosxdx}{\sqrt{1+sin^{2}x } }" align="absmiddle" class="latex-formula">

Answer 1

Substitute sin(x) = tan(t) and cos(x) dx = sec²(t) dt. We want this change of variable to be reversible, so let's assume bot x and t are bounded between 0 and π/2.

Then we have

$\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \int \frac{\sec^2(t)}{\sqrt{1 + \tan^2(t)}} \, dt$

Recall the Pythagorean identity,

1 + tan²(t) = sec²(t)

Then

√(1 + tan²(t)) = √(sec²(t)) = sec(t)

and the integral reduces to

$\displaystyle \int \frac{\sec^2(t)}{\sqrt{1 + \tan^2(t)}} \, dt = \int \frac{\sec^2(t)}{\sec(t)} \, dt = \int \sec(t) \, dt = \ln|\sec(t)+\tan(t)| + C$

Change the variable back to x, so the antiderivative is

$\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \ln \left|\sec\left(\tan^{-1}(\sin(x))\right) + \tan\left(\tan^{-1}(\sin(x))\right) \right| + C$

$\displaystyle \int \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx = \boxed{\ln \left|\sqrt{1+\sin^2(x)} + \sin(x) \right| + C}$