Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \tan x + \cot x \right)^2 d x . Then, \]
\[I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \tan^2 x + \cot^2 x + 2 \tan x \cot x \right) dx\]
\[ \Rightarrow I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \tan^2 x + \cot^2 x + 2 \right) dx\]
\[ \Rightarrow I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \sec^2 x - 1 + {cosec}^2 x - 1 + 2 \right) dx\]
\[ \Rightarrow I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \sec^2 x + {cosec}^2 x \right) dx\]
\[ \Rightarrow I = \left[ \tan x - \cot x \right]_\frac{\pi}{3}^\frac{\pi}{4} \]
\[ \Rightarrow I = \left( 1 - 1 \right) - \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right)\]
\[ \Rightarrow I = \frac{- 2}{\sqrt{3}}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
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"^(-4x) x^4 "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
