Advertisements
Advertisements
Question
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Sum
Solution
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`
shaalaa.com
Definite Integrals
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]
\[\int\limits_0^\pi \cos^5 x\ dx .\]
\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\] is equal to
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Find: `int logx/(1 + log x)^2 dx`
