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How did you change the color lol

They can ask everything here with build up only (or make it a last question like the 2014 hsc).

Would you continue doing math after med?

Anyone wanna try my one lol

He assumed sin^8 is a constant so it's basically integrating \sqrt{1-ax} where a is a constant

loooool, thats big brain

Ok anyone want to try this (stumped me lol): \int \frac{\sqrt{\sin{\sqrt{x}}}\cos{\sqrt{x}}}{1+x^2} dx

Second what Yonora says, there a probably much more capable people who can offer advice for free than those who offer advice for money.

So it's jsut using the result \int \frac{f'(x)}{1+(f(x))^2} dx =\tan^{-1}(f(x))+C If you want you could do u=\cos(x) and proceed from there.

Or we can use the king Rule: I=\int_{0}^{\pi} \frac{2x \sin{x}}{3+\cos{2x}} dx (1) I=-\int_{\pi}^{0} \frac{2 (\pi-x) \sin{x}}{3+\cos{2x}} dx \quad \text{by letting} \quad u=\pi-x I=\int_{0}^{\pi} \frac{2 (\pi-x) \sin{x}}{3+\cos{2x}} dx (2) Add 1 and 2 to get: I=\int_{0}^{\pi}...

The answer is: \frac{{\pi}^2}{4}

Thats as simple as it gets I think (you can usually simplify if the inverse is in the brackets) also try using \cos^{-1}(x) to get \cos^{-1}(x) imo looks a lot better

wtf is this account

Super small typing error here.

Really cool integral from stanford which a lot of people have probably seen before: \int_{0}^{\pi} \frac{2x \sin{x}}{3+\cos{2x}} dx

Same adding and subtracting the same thing trick as the previous question: \int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx =\int\frac{x^2\tan^{-1}x}{1+x^2}dx =\int\frac{(x^2+1-1)\tan^{-1}x}{1+x^2}dx =\int \tan^{-1}(x)-\frac{\tan^{-1}x}{1+x^2}dx IBP on the first integral gives...

Let x=\sin{u} \implies dx=\cos{u} du \int x dx =\int \sin{u} \cos{u} du =\frac{1}{2}\int \sin{2u} du =-\frac{1}{4} \cos{2u}+C where C is a constant =-\frac{1}{4} (1-2\sin^2{x})+C =-\frac{1}{4} (1-2x^2)+C =-\frac{1}{4} +\frac{1}{2}x^2+C =\frac{1}{2}x^2+C_1 where C_1 is another constant

\int \frac{1}{(x+a)^\frac{8}{7}(x-b)^\frac{6}{7}} dx where a and b are constants

\int \frac{1}{x+x^6}dx =\int \frac{1}{x^6(\frac{1}{x^5}+1)} dx =-\frac{1}{5}\ln \left| \frac{1}{x^5}+1 \right|+C OR \int \frac{1}{x+x^6}dx =\int \frac{1+6x^5-6x^5}{x+x^6}dx =\int \frac{1+6x^5}{x+x^6}-\frac{6x^4}{1+x^5}dx =\ln |x+x^6|-\frac{6}{5}\ln \left|1+x^5\right| +C

For part d: R.T.P \frac{F_{2k}}{F_{2k-1}}<\frac{F_{2k+2}}{F_{2k+1}} for positive k. I will use Binet's formula again (I feel like I'm abusing it at this point lol): F_{2k+2}F_{2k-1}-F_{2k}F_{2k+1} =\frac{(\phi^{2k+2}-\psi^{2k+2})}{\sqrt{5}} \cdot...