What is the integral of #int cos^2(x) tan^3(x) dx#?

Question

Ezekiel Alexander · Accepted Answer

#((tanx)^2/2)+c# #int(cosx)^2(tanx)^3(dx)# =#int(cosx)^2times(sinx)^3/(cosx)^3times(dx)# =#int(sinx)^3/(cosx)(dx)# =#int(tanx)(secx)^2(dx)# Let #tanx = t# #:.sec^2x dx = dt# Replacing, =#inttdt# =#t^2/2+c# Replacing back, =#((tanx)^2/2)+c#

Emma Aldridge · Answer

undefined The integral of $ \int \cos^2(x) \tan^3(x) \, dx $ can be solved using trigonometric identities. By applying the identity $ \tan^2(x) = \sec^2(x) - 1 $, and then expressing $ \cos^2(x) $ in terms of $ \sec^2(x) $, the integral can be rewritten in a form that can be more easily integrated.

After applying the trigonometric identities, the integral simplifies to $ \int (\sec^2(x) - 1) \tan^3(x) \, dx $.

Now, let $ u = \tan(x) $, then $ du = \sec^2(x) \, dx $.

Substituting these into the integral, we get $ \int (u^3 - u) \, du $.

Integrating term by term, we obtain $ \frac{1}{4}u^4 - \frac{1}{2}u^2 + C $, where $ C $ is the constant of integration.

Finally, substituting back $ u = \tan(x) $, the integral becomes $ \frac{1}{4}\tan^4(x) - \frac{1}{2}\tan^2(x) + C $, where $ C $ is the constant of integration.