Advertisements
Advertisements
Question
Advertisements
Solution
\[\int_0^\frac{\pi}{2} \cos^2 x\ d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 + \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 + \cos2x \right) dx\]
\[ = \frac{1}{2} \left[ x + \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left[ \frac{\pi}{2} + 0 \right]\]
\[ = \frac{\pi}{4}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
