Today's calculation of Integral 495

kunny

Evaluate the following definite integrals.

(1) $\int_0^{\frac {1}{2}} \frac {x^2}{\sqrt {1 - x^2}}\ dx$

(2) $\int_0^1 \frac {1 - x}{(1 + x^2)^2}\ dx$

(3) $\int_{ - 1}^7 \frac {dx}{1 + \sqrt [3]{1 + x}}$

earldbest

are the answers

(1) $\frac {\pi}{12} - \frac {1}{8}$ ?

(2) [edit] $-\frac {73}{60}$ ?

.... and are there really two $dx$ 's in number (3)? Sad

_________________
Ninja

42 is the answer to the universe.

kunny

Regretabbly your answers are incorrect.

How do you find $\int_0^{\frac {1}{2}} \frac {1}{\sqrt {1 - x^2}}\ dx$ without using $\sin ^{ - 1}x$ ?

**Posted:** Thu Oct 08, 2009 8:24 pm

earldbest

**Posted:** Fri Oct 09, 2009 2:08 am

_________________
Ninja

42 is the answer to the universe.

kunny

Still incorrect. Sad

**Posted:** Fri Oct 09, 2009 5:16 am

makar

**Posted:** Fri Oct 09, 2009 9:59 am

kunny

That's correct. Very Happy

**Posted:** Fri Oct 09, 2009 10:04 am

makar

**Posted:** Fri Oct 09, 2009 10:14 am

Dr Sonnhard Graubner

hello, for (3), setting $\sqrt [3]{1 + x} = t$ we get $dx = 3t^2\,dt$ and our integral is
$3\int_{0}^{2}\frac {t^2}{1 + t}\,dt = 3[\frac {1}{2}t^2 - t + \ln(1 + t)]_{0}^{2} = 3\ln(3)$
Sonnhard.