Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int\limits_0^{\pi/2} \sin^2 x\ dx .\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 - \cos2x \right) dx\]
\[ = \frac{1}{2} \left[ x - \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
Write the coefficient a, b, c of which the value of the integral
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. 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 ______.
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
