Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
Advertisements
उत्तर
\[\int_0^\frac{\pi}{2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos^2 x}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos x}{1 + \cos x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos x - 1 + 1}{\left( 1 + \cos x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{2 - \left( 1 + \cos x \right)}{\left( 1 + \cos x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{2}{1 + \cos x}dx - \int_0^\frac{\pi}{2} dx\]
\[ = \int_0^\frac{\pi}{2} \frac{2\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}dx - \int_0^\frac{\pi}{2} dx\]
\[ = 2 \int_0^\frac{\pi}{2} \frac{1 - \cos x}{\sin^2 x}dx - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 \int_0^\frac{\pi}{2} \left( \ cosec^2 x - \ cosec x\ cotx \right) dx - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 \left[ - cotx + \ cosec x \right]_0^\frac{\pi}{2} - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 - \frac{\pi}{2}\]
APPEARS IN
संबंधित प्रश्न
If f is an integrable function, show that
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
