Advertisements
Advertisements
प्रश्न
\[\int\limits_0^1 \tan^{- 1} x dx\]
उत्तर
\[\int_0^1 \tan^{- 1} x d x\]
\[ = \int_0^1 \tan^{- 1} x \times 1 d x\]
\[ = \left[ \tan^{- 1} x x \right]_0^1 - \int_0^1 \frac{x}{1 + x^2}dx\]
\[ = \left[ x \tan^{- 1} x \right]_0^1 - \frac{1}{2} \left[ \log\left( 1 + x^2 \right) \right]_0^1 \]
\[ = \frac{\pi}{4} - 0 - \frac{1}{2}\log2 + 0\]
\[ = \frac{\pi}{4} - \frac{1}{2}\log2\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
Γ(1) is
