Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\pi \cos^5 x d x\]
\[ = \int_0^\pi \cos x \left( \cos^2 x \right)^2 dx\]
\[ = \int_0^\pi \cos x \left( 1 - \sin^2 x \right)^2 dx\]
\[ Let \sin x = t, then\ \cos x\ dx = dt\]
\[When\, x \to 0 ; t \to 0\ and\ x \to \pi ; t \to 0\]
\[ \text{Therefore}, \]
\[I = \int_0^0 \left( 1 - t^2 \right)^2 dt\]
\[ = 0\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
Evaluate each of the following integral:
Write the coefficient a, b, c of which the value of the integral
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
