Advertisements
Advertisements
Question
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
Advertisements
Solution
\[Let, I = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^3 x} d x ..............(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^3 \left( \frac{\pi}{2} - x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^3 x} d x ................(2)\]
Adding (1) and (2)
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{1}{1 + \tan^3 x} + \frac{1}{1 + co t^3 x} \right] d x\]
\[ = \int_0^\frac{\pi}{2} \frac{2 + \tan^3 x + co t^3 x}{\left( 1 + \tan^3 x \right)\left( 1 + co t^3 x \right)}dx\]
\[ = \int_0^\frac{\pi}{2} \frac{2 + \tan^3 x + co t^3 x}{2 + \tan^3 x + co t^3 x}dx\]
\[ = \int_0^\frac{\pi}{2} dx \]
\[ = \left( x \right)_0^\frac{\pi}{2} \]
\[ = \frac{\pi}{2}\]
\[Hence, I = \frac{\pi}{4}\]
APPEARS IN
RELATED QUESTIONS
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Evaluate each of the following integral:
Evaluate each of the following integral:
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following:
Γ(4)
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
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` = ______.
