Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \cos^4 x\ dx\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \left( \cos^2 x \right)^2 dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\left( 1 + \cos 2x \right)^2}{4} dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( 1 + \cos^2 2x + 2 \cos 2x \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( 1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \int_0^\frac{\pi}{2} \left( \frac{3 + 4 \cos 2x + \cos 4x}{2} \right) dx\]
\[ \Rightarrow I = \frac{1}{4} \left[ \frac{3x}{2} + \frac{2 \sin 2x}{2} + \frac{\sin 4x}{8} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{1}{4}\left[ \frac{3\pi}{4} + 0 \right]\]
\[ \Rightarrow I = \frac{3\pi}{16}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
