PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 

\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\]  equals

\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals

\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`

Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]

The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals

The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is