Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
Advertisements
उत्तर
\[\int_0^\frac{\pi}{2} \frac{\cos x}{1 + \sin^2 x} d x\]
\[Let \sin x = t,\text{ then }\cos x dx = dt\]
\[\text{When }x \to 0 ; t \to 0\]
\[\text{And }x \to \frac{\pi}{2}; t \to 1\]
Therefore the integral becomes
\[ \int_0^1 \frac{dt}{1 + t^2}\]
\[ = \left[ \tan^{- 1} x \right]_0^1 \]
\[ = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
