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No it's not a 4U question lol. 4U you could get something like.. \int \frac{x}{1 + x^6} dx

:S Did not realise the edit.. Should have double checked.

Of course lol. Whenever you substitute, you have to change everything in terms of that substitution (the dx -> d(theta), the original expression -> new expression, limits -> new limits) You only need to use the triangle when going back to the original variable.

Thanks for the rep man :D

Textbooks do what you just asked for pretty well lol.

Wolframalpha tells me otherwise.. Link Either me, wolframalpha and barbenator are wrong, the Fitzpatrick book is wrong, or you typed the question wrong.

Serious stuff.

Use polynomial long division. Or you could do this.. \frac{x^2 - 5x + 2}{x - 2} = \frac{x^2 - 5x}{x - 2} + \frac{2}{x - 2} = \frac{x(x - 2) - 3x}{x - 2} + \frac{2}{x - 2} = x - \frac{3x - 6}{x - 2} - \frac{6}{x - 2} + \frac{2}{x - 2} = x - 3 - \frac{4}{x - 2}

LOL. Can't wait.

I've always thought lower case delta was used for smaller changes and upper case delta for larger changes lol.

\int_{0}^{\frac{a}{2}} x^{2}\sqrt{a^{2}-x^{2}}dx $Let $ x = asin\theta dx = acos\theta d\theta x = 0, \theta = 0 x = \frac{a}{2}, \theta = \frac{\pi}{6} \therefore \int a^2sin^2\theta \sqrt{a^2 - a^2sin^2\theta} acos\theta d\theta = \int a^4 sin^2\theta cos^2\theta d\theta = \frac{a^4}{4}...

I believe he meant to type arctany, then differentiated that.

\frac{d}{d\theta}(tan\theta) = sec^2\theta $Don't believe me?$ \frac{d}{d\theta}(tan\theta) = \frac{d}{d\theta}(\frac{sin\theta}{cos\theta}) = \frac{cos^2\theta + sin^2\theta}{cos^2\theta} = \frac{1}{cos^2\theta} = sec^2\theta \therefore \int \frac{d}{d\theta}(tan\theta) d\theta = \int...

1. Use double angle formula to change sin^2x in terms of cos2x Use t-substitution 2. Multiply top and bottom by (1 - sinx), split the integral and you'll end up with something nice. Edit: lol. everyone posts too quickly.

Updated post.

3 was fun. :D I used simultaneous equations (had to force myself to remember little rules i set so that i don't get them mixed up) and i ended up with 72. $Let the integer be XY where X and Y are positive integers between 1 and 10 where $ 10 < XY < 100 XY = 8(X + Y) XY = 8X + 8Y 10X = 8X +...

Varies. There's no set number.

Questions sorted by section (1, 2 or 3).

That gives the voltage of each individual cell (~2.1V). There are generally 6 cells so multiply that by 6 to get your total voltage of around 12.6V

Seems right. I don't think i've ever drawn anything as complicated as that back in year 11. All i remember drawing are NaCl, HCl, water and simple diatomic molecules haha.