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
Solve each of the following integral:
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
Find: `int logx/(1 + log x)^2 dx`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
