exo7 4267

$\int_{t=0} ^{\pi/2} \cos^4t \,d t = \frac{3\pi}{16}$

$\int_{t=-\pi/2} ^{\pi/2} \sin^2t\cos^3t \,d t = \frac4{15}$

$\int_{t=0} ^{\pi/2} t^2\cos t \,d t = \frac{\pi^2}4 - 2$

$\int_{t=-\pi/2} ^{\pi/2} t^2\sin t\cos^2t \,d t = 0$

$\int_{t=0} ^{\pi/2} \frac{\sin t}{1+\cos^2t} \,d t = \frac \pi4$

$\int_{t=0} ^{\pi/2} \frac{d t}{1 + \sin t} = 1$

$\int_{t=0} ^{\pi/2} \frac{\sin^2t}{\sin t + \cos t} \,d t = -\frac1{\sqrt2} \ln(\sqrt2-1)$

$\int_{t=0} ^{\pi/2} \frac{\sin2t}{\sqrt{1-a\sin t}} \,d t = \frac{4(2-(a+2)\sqrt{1-a})}{3a^2}$

$\int_{t=0} ^1 t\ln t \,d t = -\frac14$

$\int_{t=0} ^1 \Arcsin t \,d t = \frac\pi2 - 1$

$\int_{t=0} ^3 \frac{2t}{(1+t^2)(3+t^2)} \,d t = \frac12 \ln\frac52$

$\int_{t=0} ^1 \frac{t^2\Arctan t}{1+t^2} \,d t = \frac\pi4 - \frac{\pi^2}{32} - \ln\sqrt2$

$\int_{t=0} ^{\ln2} \sqrt{e^t-1} \,d t = 2-\frac\pi2$

$\int_{t=4} ^9 \frac{d t}{\sqrt t - 1} = 2 + 2\ln2$

$\int_{t=0} ^1 \frac{te^t}{\sqrt{e^t+1}} \,d t = 4\sqrt2 - 2\sqrt{e+1} + 4\ln\Bigl[ \frac{\sqrt{e+1}+1}{\sqrt2+1} \Bigr] - 2$

$\int_{t=0} ^1 \frac{\ln(1-a^2t^2)}{t^2} \,d t = a\ln\left|\frac{1-a}{1+a}\right| -\ln(1-a^2)$

$\int_{t=0} ^1 \frac{d t}{2+\sqrt{1-t^2}} = \frac\pi6 (3 - \frac4{\sqrt3})$

$\int_{t=-1} ^1 \frac{d t}{t+\sqrt{t^2+1}} = \ln(1+\sqrt2) + \sqrt2$

$\int_{t=-1} ^1 \sqrt{1+t^2} \,d t = \ln(1+\sqrt2) + \sqrt2$