Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
Advertisements
उत्तर
\[Let, I = \int_0^\frac{\pi}{2} \frac{x}{\sin^2 x + \cos^2 x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{x}{1} d x\]
\[ = \int_0^\frac{\pi}{2} x d x\]
\[ = \left[ \frac{x^2}{2} \right]_0^\frac{\pi}{2} \]
\[ \therefore I = \frac{\pi^2}{8}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] 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
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
