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
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
