Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^1 x \tan^{- 1} x\ d\ x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \int_0^1 \frac{x^2}{1 + x^2} dx\]
\[ \Rightarrow I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \int_0^1 \left( \frac{1 + x^2}{1 + x^2} - \frac{1}{1 + x^2} \right) dx\]
\[ \Rightarrow I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \left[ x - \tan^{- 1} x \right]_0^1 \]
\[ \Rightarrow I = \frac{\pi}{8} - 0 - \frac{1}{2}\left( 1 - \frac{\pi}{4} - 0 \right)\]
\[ \Rightarrow I = \frac{\pi}{4} - \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
