Advertisements
Advertisements
प्रश्न
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
पर्याय
0
2
1
4
MCQ
Advertisements
उत्तर
2
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^2 x\sqrt{x + 2}\ dx\]
\[\int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]
\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]
\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]
If f is an integrable function, show that
\[\int\limits_{- a}^a x f\left( x^2 \right) dx = 0\]
\[\int\limits_0^\infty e^{- x} dx .\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx\] equals to
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
