Basic integrations and evaluations (calculus 2 work)

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Homework Statement



I have a few questions, and I already have (some of) the right answers (or so I think.. but I don't understand them)

(Just finding the anti-derivative)
1. http://www.codecogs.com/eq.latex?\int_{}^{}\sin\left(8z-5\%20\right)
My answer: -cos (4z^2-5z) (not sure if I have to follow the chain rule in this or not?) If not I think this is right.

(Evaluation)
2. http://www.codecogs.com/eq.latex?\int_{1}^{0}3x^{2}+x-5
My answer: 3.5

When I evaluated this I had
-1 + -1/2 + 5

(Evaluation)
3. http://www.codecogs.com/eq.latex?\int_{-\sqrt[]{3}}^{\sqrt[]{3}}(t+1)(t^{2}+4)
My answer: 29.32
My anti-derivative: 1/4t^4 + 1/3t^3 + 2t^2 + 4t

(Evaluation)
4. http://www.codecogs.com/eq.latex?\int_{1/2}^{1}(1/u^{3}%20-%201/u^{4})
My answer: .833
Antiderivative: -1/2v^-2 - -1/3v^-3


Am I doing these right?
 
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Oh, I forgot to mention.

On #2 and #4 someone put them into some computer software and said they came out negative... but I can't find any errors on my work?
 
Caveat: I looked at #2 and #4 only.

Your answer for #2 is correct, based on the limits of integration you showed. Your friend might have put the limits of integration into the application in the opposite order, which explains the sign discrepancy.

Your answer for #4 is the right number, but the wrong sign. I get a value of -5/6.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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