Advertisements
Advertisements
Question
Advertisements
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
Prove that:
Evaluate each of the following integral:
Evaluate each of the following integral:
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
What does a definite integral represent?
