If f(x) = int_0^{pi/2} frac{ln(1 + x sin^2 th | vedclass

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

Question

(D) Let $I(x) = \int_0^{\pi/2} \frac{\ln(1 + x \sin^2 	heta)}{\sin^2 	heta} d	heta$. Using Leibniz rule for differentiation under the integral sign

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

Vedclass · Answer

(D) Let $I(x) = \int_0^{\pi/2} \frac{\ln(1 + x \sin^2 	heta)}{\sin^2 	heta} d	heta$. Using Leibniz rule for differentiation under the integral sign: $f'(x) = \int_0^{\pi/2} \frac{\partial}{\partial x} \left( \frac{\ln(1 + x \sin^2 	heta)}{\sin^2 	heta} ight) d	heta = \int_0^{\pi/2} \frac{1}{1 + x \sin^2 	heta} d	heta$. Multiply numerator and denominator by $\sec^2 	heta$: $f'(x) = \int_0^{\pi/2} \frac{\sec^2 	heta}{\sec^2 	heta + x 	an^2 	heta} d	heta = \int_0^{\pi/2} \frac{\sec^2 	heta}{1 + (1+x) 	an^2 	heta} d	heta$. Let $u = \sqrt{1+x} 	an 	heta$,then $du = \sqrt{1+x} \sec^2 	heta d	heta$. $f'(x) = \frac{1}{\sqrt{1+x}} \int_0^{\infty} \frac{du}{1+u^2} = \frac{1}{\sqrt{1+x}} [	an^{-1} u]_0^{\infty} = \frac{\pi}{2\sqrt{1+x}}$. Integrating $f'(x)$ with respect to $x$: $f(x) = \int \frac{\pi}{2\sqrt{1+x}} dx = \pi \sqrt{1+x} + C$. Since $f(0) = \int_0^{\pi/2} \frac{\ln(1)}{\sin^2 	heta} d	heta = 0$,we have $\pi \sqrt{1} + C = 0 \implies C = -\pi$. Thus,$f(x) = \pi(\sqrt{x+1} - 1)$. Both $(A)$ and $(B)$ are correct.

If $f(x) = \int_0^{\pi/2} \frac{\ln(1 + x \sin^2 \theta)}{\sin^2 \theta} d\theta$,$x \geq 0$,then:

Similar Questions

Let $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$,then:

$\int_{-5}^5 x^4\left(25-x^2\right)^{5 / 2} d x=$

Let $F(x) = \int_x^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt$ for all $x \in \mathbb{R}$ and $f: [0, \frac{1}{2}] \rightarrow [0, \infty)$ be a continuous function. For $a \in [0, \frac{1}{2}]$,if $F'(a) + 2$ is the area of the region bounded by $x=0, y=0, y=f(x)$ and $x=a$,then $f(0)$ is

Let $S(x) = \int_{x^2}^{x^3} \ln t \, dt$ for $x > 0$ and $H(x) = \frac{S'(x)}{x}$. Then $H(x)$ is :

If $\int f(x) dx = F(x) + C$,then $\frac{d}{dt} \int_{g(t)}^{h(t)} f(x) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module