Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ e^x = t . Then, e^x dx = dt\]
\[When\ x = 0, t = 1\ and\ x = 1, t = e\]
\[ \therefore I = \int_0^1 \frac{e^x}{1 + e^{2x}} d x\]
\[ \Rightarrow I = \int_1^e \frac{dt}{1 + t^2}\]
\[ \Rightarrow I = \left[ \tan^{- 1} x \right]_1^e \]
\[ \Rightarrow I = \tan^{- 1} e - \tan^{- 1} 1\]
\[ \Rightarrow I = \tan^{- 1} e - \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate :
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following:
`Γ (9/2)`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
