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 each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
