Advertisements
Advertisements
प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Advertisements
उत्तर
\[Let\ I = \int_\frac{\pi}{4}^\frac{\pi}{2} \cot x\ d\ x\ . Then, \]
\[I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \cot x\frac{- (cosec x + \cot x)}{cosec x + \cot x} dx\]
\[ \Rightarrow I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{- cosec x \cot x - \cot^2 x}{cosec x + \cot x} dx\]
\[ \Rightarrow I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{- cosec x \cot x - {cosec}^2 x + 1}{cosec x + \cot x} dx \left[ \because {cosec}^2 x = 1 + \cot^2 x \right]\]
\[ \Rightarrow I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{- cosec x \cot x - {cosec}^2 x}{cosec x + \cot x} dx - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{1}{cosec x + \cot x}dx\]
\[ \Rightarrow I = - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{- cosec x \cot x - {cosec}^2 x}{cosec x + \cot x} dx - \int_\frac{\pi}{4}^\frac{\pi}{2} \frac{\sin x}{1 + \cos x}dx\]
\[ \Rightarrow I = - \left[ \log \left( cosec x + \cot x \right) \right]_\frac{\pi}{4}^\frac{\pi}{2} + \left[ \log \left( 1 + \cos x \right) \right]_\frac{\pi}{4}^\frac{\pi}{2} \]
\[ \Rightarrow I = - \log \left( 1 + \infty \right) + \log \left( \sqrt{2} + 1 \right) + \log \left( 1 + 0 \right) - \log \left( 1 + \frac{1}{\sqrt{2}} \right)\]
\[ \Rightarrow I = \log \left( \sqrt{2} + 1 \right) - \log \left( \frac{\sqrt{2} + 1}{\sqrt{2}} \right)\]
\[ \Rightarrow I = \log \left( \frac{\sqrt{2}\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} + 1 \right)} \right)\]
\[ \Rightarrow I = \log\sqrt{2}\]
\[ \Rightarrow I = \frac{1}{2}\log 2\]
APPEARS IN
संबंधित प्रश्न
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Evaluate :
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
