106for Q3. Let x-α = t2
So we have ∫2tdtt*(β-α-t2)0.5
now it is done
12 0∫1 log~+[x] = 0∫1 log~
because [x] =0 for x in [0,1), so it wont affect the area under the curve......
The remaining integral is simple for everyone else, I mean, I am horrible at integrals
A mental graph, shows that the value is between log√2 and log2
21\int_{0}^{\pi /2}csc(x-\frac{\pi}{3}).csc(x-\frac{\pi}{6})dx=2\int_{0}^{\pi /2}\frac{sin\left ( (x-\pi/6)-(x-\pi/3) \right )}{sin(x-\frac{\pi}{3}).sin(x-\frac{\pi}{6})} dx=2\int_{0}^{\pi /2}\frac{sin\left(x-\pi/6)cos(x-\pi/3)-cos(x-\pi/6)sin(x-\pi/3) }{sin(x-\frac{\pi}{3}).sin(x-\frac{\pi}{6})} dx=\\\\\2\int_{0}^{\pi/2}\left ( cot(x-\pi/3) -cot(x-\pi/6) \right dx =2\left ( \left \left | log\left | sin(x-\pi/3) \right | +log\left | sin(x-\pi/6) \right |\right \right |^{\pi/2} _{0} \right )
1\int_{0}^{\frac{\pi}{2}}{\frac{dx}{\sin\left( {x-\frac{\pi}{3}}\right).\sin\left( {x-\frac{\pi}{6}}\right)}} \\ \\ \\ \frac{1}{\sin\left( \left(x-\frac{\pi}{3} \right)-\left(x-\frac{\pi}{6} \right)\right)}\int_{0}^{\frac{\pi}{2}}{\frac{\sin\left( \left(x-\frac{\pi}{3} \right)-\left(x-\frac{\pi}{6} \right)\right)dx}{\sin\left( {x-\frac{\pi}{3}}\right).\sin\left( {x-\frac{\pi}{6}}\right)}}\\ \\\texttt{now expand using sin(A-B) formula }
21btw for 2nd one u can do it firstly by parts takin log~ as 1st fuction and 1 as the second then put teh limits..
11For the 2nd one...
2I=\int_{0}^{1}{ln(2+2\sqrt{1-x^2})}dx
From which we just need the integral \int_{0}^{\frac{\pi}{2}}{ln\left( 2cos\frac{a}{2}\right)cosada}
From which we just need the evaluation of \int_{0}^{\frac{\pi}{2}}{ln\left( cos\frac{a}{2}\right)cosada}.....Byparts does the rest.
11The last integral is very interesting.....
Putting x=\alpha cos^2\gamma+\beta sin^2\gamma gives the answer as \pi....however a different approach gives a different result.
I=\int_{\alpha}^{\beta}{\sqrt{\frac{x-\alpha}{x-\beta}}}dx-\int_{\alpha}^{\beta}{\sqrt{\frac{\beta-x}{x-\alpha}}dx
Applying \int_{a}^{b}{f(x)dx}=\int_{a}^{b}{f(a+b-x)dx}, we have \boxed{I=0}....which is correct?
