Question

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sure where to go from there. U-sub won't work and neither will taking out a tan(pheta). Please help. Thanks

Answer 1

Setting , you have . Then the integral becomes






Now, in general. But since we want our substitution to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means , which implies that , or equivalently that . Over this domain, , so .

Long story short, this allows us to go from



to




Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get



Then integrate term-by-term to get




Now undo the substitution to get the antiderivative back in terms of .



and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

Answer 2

`\displaystyle\int\frac{√(25-x^2)}x\,\mathrm dx`

Setting
`x=5\sin\theta`, you have
`\mathrm dx=5\cos\theta\,\mathrm d\theta`. Then the integral becomes


`\displaystyle\int(√(25-(5\sin\theta)^2))/(5\sin\theta)5\cos\theta\,\mathrm d\theta`

`\displaystyle\int√(25-25\sin^2\theta)(\cos\theta)/(\sin\theta)\,\mathrm d\theta`

`\displaystyle5\int√(1-\sin^2\theta)(\cos\theta)/(\sin\theta)\,\mathrm d\theta`

`\displaystyle5\int√(\cos^2\theta)(\cos\theta)/(\sin\theta)\,\mathrm d\theta`

Now,
`√(x^2)=|x|` in general. But since we want our substitution
`x=5\sin\theta` to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means
`\theta=\sin^(-1)\frac x5`, which implies that
`\left|\frac x5\right|\le1`, or equivalently that
`|\theta|\le\frac\pi2`. Over this domain,
`\cos\theta\ge0`, so
`√(\cos^2\theta)=|\cos\theta|=\cos\theta`.

Long story short, this allows us to go from


`\displaystyle5\int√(\cos^2\theta)(\cos\theta)/(\sin\theta)\,\mathrm d\theta`

to


`\displaystyle5\int\cos\theta(\cos\theta)/(\sin\theta)\,\mathrm d\theta`

`\displaystyle5\int(\cos^2\theta)/(\sin\theta)\,\mathrm d\theta`

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get


`(\cos^2\theta)/(\sin\theta)=(1-\sin^2\theta)/(\sin\theta)=\csc\theta-\sin\theta`

Then integrate term-by-term to get


`\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)`

`=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C`

Now undo the substitution to get the antiderivative back in terms of
`x`.


`=-5\ln\left|\csc\left(\sin^(-1)\frac x5\right)+\cot\left(\sin^(-1)\frac x5\right)\right|+\cos\left(\sin^(-1)\frac x5\right)+C`

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to


`=-5\ln\left|\frac{5+√(25-x^2)}x\right|+√(25-x^2)+C`