Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ I = \int_0^1 \tan^{- 1} x\ d\ x\ . Then, \]
\[I = \int_0^1 1 \tan^{- 1} x\ d\ x\]
\[\text{Integrating by parts}\]
\[I = \left[ x \tan^{- 1} x \right]_0^1 - \int_0^1 \frac{x}{1 + x^2} d x\]
\[ \Rightarrow I = \left[ x \tan^{- 1} x \right]_0^1 - \frac{1}{2} \left[ \log \left( x^2 + 1 \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{\pi}{4} - 0 - \frac{1}{2} \log 2 + 0\]
\[ \Rightarrow I = \frac{\pi}{4} - \frac{1}{2} \log 2\]
APPEARS IN
संबंधित प्रश्न
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
