Advertisements
Advertisements
प्रश्न
उत्तर
\[Let I = \int_0^\frac{\pi}{2} \cos^2 x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^2 x\ d\ x\]
\[ \Rightarrow I = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 + \cos 2x \right) dx \left[ \because \cos 2x = 2 \cos^2 x - 1 \right]\]
\[ \Rightarrow I = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi}{4} + 0 - 0\]
\[ \Rightarrow I = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Prove that:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Choose the correct alternative:
`Γ(3/2)`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is: