Advertisements
Advertisements
Question
Solution
\[\text{We have}, \]
\[I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x\]
\[Let f\left( x \right) = x \cos^2 x\]
\[ \Rightarrow f\left( - x \right) = \left( - x \right) \cos^2 \left( - x \right)\]
\[ = - x \cos^2 x\]
\[ \therefore f\left( - x \right) = - f\left( x \right)\]
\[i . e . , f\left( x \right) \text{is odd function}\]
\[\text{We know that} \int_{- a}^a f\left( x \right) d x = 0 , \text{if }f\left( x \right) \text{is odd function} . \]
\[ \therefore I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x = 0\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "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
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
