Advertisements
Advertisements
प्रश्न
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
उत्तर
Let f(θ)= sin2θ
f(– θ) = sin2(– θ)
= [sin (– θ)]2
= [– sin θ]2
= sin2θ
f(– θ) = f(θ)
∴ f(θ) is an even function
`int_(- pi/2)^(pi/2) sin^2theta "d"theta = 2 xx int_0^(pi/2) sin^2theta "d"theta`
= `2 xx int_0^(pi/2) ((1 - cos 2theta)/2) "d"theta`
= `2 xx 1/2 int_0^(pi/2) (1 - cos 2theta) "d"theta`
= `[theta - (sin 2theta)/2]_0^(pi/2)`
= `[pi/2 - (sin2(pi/2))/2] - [0]`
= `pi/2 - 0`
= `pi/2`
APPEARS IN
संबंधित प्रश्न
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
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):}`