Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
Advertisements
उत्तर
\[\int_0^\frac{\pi}{4} \tan^4 x d x\]
\[ = \int_0^\frac{\pi}{4} \tan^2 x\left( se c^2 x - 1 \right) d x\]
\[ = \int_0^\frac{\pi}{4} \tan^2 x se c^2 x dx - \int_0^\frac{\pi}{4} \tan^2 x dx\]
\[ = \left[ \frac{\tan^3 x}{3} \right]_0^\frac{\pi}{4} - \left[ \tan x - x \right]_0^\frac{\pi}{4} \]
\[ = \frac{1}{3} - 1 + \frac{\pi}{4}\]
\[ = \frac{\pi}{4} - \frac{2}{3}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Choose the correct alternative:
Γ(1) is
Choose the correct alternative:
`Γ(3/2)`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
Evaluate \[\int_{0}^{1}x\,dx\].
