Advertisements
Advertisements
Question
Options
0
1
π/2
π/4
Advertisements
Solution
π/4
We have
\[ I = \int_0^\frac{\pi}{2} \frac{1}{1 + \cot^3 x} d x . . . . . \left( 1 \right)\]
\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \cot^3 \left( \frac{\pi}{2} - x \right)} d x \]
\[ \therefore I = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^3 x} d x . . . . . \left( 2 \right)\]
\[\text{Adding} \left( 1 \right) and \left( 2 \right) \text{we get}\]
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{1}{1 + co t^3 x} + \frac{1}{1 + \tan^3 x} \right] d x\]
\[= \int_0^\frac{\pi}{2} \left[ \frac{1 + \tan^3 x + 1 + co t^3 x}{\left( 1 + co t^3 x \right)\left( 1 + \tan^3 x \right)} \right] dx\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{2 + \tan^3 x + co t^3 x}{1 + \tan^3 x + co t^3 x + co t^3 x \tan^3 x} \right]dx\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{2 + \tan^3 x + co t^3 x}{1 + \tan^3 x + co t^3 x + 1} \right]dx\]
\[ = \int_0^\frac{\pi}{2} \left[ \frac{2 + \tan^3 x + co t^3 x}{2 + \tan^3 x + co t^3 x} \right] dx\]
\[ = \int_0^\frac{\pi}{2} [1]dx\]
\[ = \left[ x \right]_0^\frac{\pi}{2} \]
\[ = \frac{\pi}{2}\]
\[Hence\ I = \frac{\pi}{4}\]
APPEARS IN
RELATED QUESTIONS
Write the coefficient a, b, c of which the value of the integral
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
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")`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following:
Γ(4)
Evaluate \[\int_{0}^{1}x\,dx\].
Evaluate \[\int_{1}^{3}(2x+1)\,dx\].
