Advertisements
Advertisements
Question
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
Advertisements
Solution
\[\int_\frac{- \pi}{2}^\frac{\pi}{2} \sin^9 x d x\]
\[\text{Let }f(x) = \sin^9 x\]
\[\text{Consider, }f(-x) = \sin^9 \left( - x \right) = - \sin^9 x = - f\left( x \right)\]
Thus f(x) is an odd function
Therefore,
\[ \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin^9 x d x = 0\]
APPEARS IN
RELATED QUESTIONS
If f(2a − x) = −f(x), prove that
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
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
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
`int_0^(2a)f(x)dx`
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
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
