Advertisements
Advertisements
Question
Advertisements
Solution
\[Let I = \int_0^\frac{\pi}{2} \cos^2 x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^2 x\ d\ x\]
\[ \Rightarrow I = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 + \cos 2x \right) dx \left[ \because \cos 2x = 2 \cos^2 x - 1 \right]\]
\[ \Rightarrow I = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi}{4} + 0 - 0\]
\[ \Rightarrow I = \frac{\pi}{4}\]
APPEARS IN
RELATED QUESTIONS
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
