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प्रश्न
विकल्प
- \[\frac{\pi}{12}\]
- \[\frac{\pi}{6}\]
- \[\frac{\pi}{4}\]
- \[\frac{\pi}{3}\]
\[\frac{\pi}{2}\]
\[\frac{2\pi}{3}\]
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उत्तर
Explanation:
Let I = `int_1^sqrt(3) dx/(1 + x^2)`
= `[tan^-1 x]_1^sqrt(3)`
= `tan^-1 sqrt(3) - tan^-1 1`
= `π/3 - π/4`
= `(4π - 3π)/12`
= `π/12`
संबंधित प्रश्न
Solve each of the following integral:
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
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_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
