How Do You Integrate \(\cos^2(x) \tan^3(x) dx\)?

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To integrate \(\int \cos^2(x) \tan^3(x) \, dx\), the discussion highlights that substitution and integration by parts may not yield success. One suggested approach involves rewriting \(\tan^3(x)\) in terms of sine and cosine, leading to \(\int \frac{\sin^3(x)}{\cos(x)} \, dx\). This can be further simplified by separating the integral into two parts: \(\int \tan(x) \, dx\) and \(\int \sin(x) \, d(\sin(x))\). The conversation concludes with a participant expressing understanding of the method presented.
Stevecgz
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I'm attempting to find:
\int \cos^2x\tan^3x dx
I've tried substitution and integration by parts but I'm having no success. Can someone help me out on how I get this one started?

Steve
 
Last edited:
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It's no big deal.

\int \frac{\sin^{3}x}{\cos x} \ dx =\int \frac{\sin x \left(1-\cos^{2}x\right)}{\cos x} \ dx =\int \tan x \ dx -\int \sin x \ d\left(\sin x\right) = ...

Daniel.
 
Thanks dextercioby, I understand it now.

Steve
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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