Evaluate $\displaystyle \int \dfrac{cos^4(\theta)}{sin(\theta)} \, d\theta$
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Since $\displaystyle \int\frac{\cos^4 t}{\sin t}dt = \int\frac{\sin t\cos^4 t}{1-\cos^2t}dt$, use the substitution $\displaystyle x =\cos t$.
What I might do is $\displaystyle \int \dfrac{\cos ^4 x}{\sin x}\,dx = \int \dfrac{\left(1 - \sin^2 x\right)^2}{\sin x}\,dx $$\displaystyle = \int \left(\csc x - 2 \sin x + \sin^3x\right)\,dx.$ It might be a little easier.
Originally Posted by Danny What I might do is $\displaystyle \int \dfrac{\cos ^4 x}{\sin x}\,dx = \int \dfrac{\left(1 - \sin^2 x\right)^2}{\sin x}\,dx $$\displaystyle = \int \left(\csc x - 2 \sin x + \sin^3x\right)\,dx.$ It might be a little easier. If you can remember off hand the antiderivatives of $\displaystyle \displaystyle \csc{x}$ and $\displaystyle \displaystyle \sin^3{x}$ (I know I can't ><, but that's just because I prefer to memorise procedures instead of integrals).
Yes, but Danny's approach is for experienced people, since remaining integrals, at least, are easy. However girdav's solution is also good, and it only uses a substitution.
Originally Posted by Krizalid Yes, but Danny's approach is for experienced people, since remaining integrals, at least, are easy. However girdav's solution is also good, and it only uses a substitution. Memorising integrals =/= Experience
Nope, that's not my point. I was talking about practice, and it does have to do with intelligence.
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