Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos^2 x\ d x\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1 + \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 1 + \cos2x \right) dx\]
\[ = \frac{1}{2} \left[ x + \frac{\sin2x}{2} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} + 0 + \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{2}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`Γ(3/2)`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
Find: `int logx/(1 + log x)^2 dx`
What does a definite integral represent?
What are definite integrals used to find over a fixed interval?
