intlimits_0^{frac{pi }{2}} frac{dx}{1 + a^2 s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int\limits_0^{\frac{\pi}{2}} \frac{dx}{1 + a^2 \sin^2 x}$.Divide numerator and denominator by $\cos^2 x$:$I = \int\limits_0^{\frac{\pi}{

Vedclass · Accepted Answer

$\frac{\pi}{2\sqrt{1 + a^2}}$

Vedclass · Answer

(A) Let $I = \int\limits_0^{\frac{\pi}{2}} \frac{dx}{1 + a^2 \sin^2 x}$.Divide numerator and denominator by $\cos^2 x$:$I = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{\sec^2 x + a^2 	an^2 x} = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + 	an^2 x + a^2 	an^2 x} = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + (1 + a^2) 	an^2 x}$.Let $u = 	an x$,then $du = \sec^2 x \, dx$. As $x 	o 0, u 	o 0$ and as $x 	o \frac{\pi}{2}, u 	o \infty$.$I = \int\limits_0^{\infty} \frac{du}{1 + (\sqrt{1 + a^2} u)^2}$.Using the formula $\int \frac{dx}{1 + k^2 x^2} = \frac{1}{k} 	an^{-1}(kx)$:$I = \left[ \frac{1}{\sqrt{1 + a^2}} 	an^{-1}(\sqrt{1 + a^2} u) ight]_0^{\infty} = \frac{1}{\sqrt{1 + a^2}} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi}{2\sqrt{1 + a^2}}$.

$\int\limits_0^{\frac{\pi }{2}} \frac{dx}{1 + a^2 \sin^2 x}$ has the value :

Similar Questions

Let $f(x) = |x - 2|$ and $g(x) = f(f(x))$,$x \in [0, 4]$. Then $\int_{0}^{3} (g(x) - f(x)) \, dx$ is equal to

If $f(x) = \begin{cases} 4x + 3, & 1 \le x \le 2 \\ 3x + 5, & 2 < x \le 4 \end{cases}$,then $\int_1^4 f(x) \, dx = $

The value of $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} dx$ is

If $\left( \int_{0}^{a} x \, dx \right) \le (a + 4)$,then

Let $\phi (x) = \int_{0}^{1} e^{x} e^{t} \phi (t) dt + x$. If $\phi (\ln (e^{2} - 3))$ is equal to $A$,then find the value of $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module