Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^\frac{\pi}{2} \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} d\ x . Then, \]
\[\text{Dividing the numerator and denominator by} \cos^2 x, we\ get\]
\[I = \int_0^\frac{\pi}{2} \frac{\sec^2 x}{a^2 \tan^2 x + b^2} d x\]
\[Let\ \tan x = t . Then, \sec^2 x\ dx\ = dt\]
\[When\ x = 0, t = 0\ and\ x\ = \frac{\pi}{2} , t = \infty \]
\[ \therefore I = \int_0^\infty \frac{1}{a^2 t^2 + b^2} d t\]
\[ \Rightarrow I = \frac{1}{a^2} \int_0^\infty \frac{1}{t^2 + \frac{b^2}{a^2}} dt\]
\[ \Rightarrow I = \frac{1}{a^2} \times \frac{a}{b} \left[ \tan^{- 1} \frac{at}{b} \right]_0^\infty \]
\[ \Rightarrow I = \frac{1}{ab}\frac{\pi}{2}\]
\[ \Rightarrow I = \frac{\pi}{2ab}\]
APPEARS IN
RELATED QUESTIONS
If f is an integrable function, show that
If f(x) is a continuous function defined on [−a, a], then prove that
Evaluate each of the following integral:
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
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_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Find `int sqrt(10 - 4x + 4x^2) "d"x`
