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प्रश्न
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उत्तर
\[\int_0^4 \frac{1}{\sqrt{16 - x^2}} d x\]
\[ = \int_0^4 \frac{1}{\sqrt{4^2 - x^2}} d x\]
\[ = \left[ \sin^{- 1} \frac{x}{4} \right]_0^4 \]
\[ = \left( \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{2}\]
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संबंधित प्रश्न
Evaluate each of the following integral:
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "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:
`int_0^oo "e"^(-mx) x^6 "d"x`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
