If x satisfies the equation left( int_{0}^{1} | vedclass

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

Question

(D) Let $I_1 = \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt$. Since $f(t) = \frac{t^2 \sin 2t}{t^2 + 1}$ is an odd function (because $f(-t) = -f(t)$)

Vedclass · Accepted Answer

$\pm 2 \sqrt{\frac{\sin \alpha}{\alpha}}$

Vedclass · Answer

(D) Let $I_1 = \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt$. Since $f(t) = \frac{t^2 \sin 2t}{t^2 + 1}$ is an odd function (because $f(-t) = -f(t)$) and the interval $[-3, 3]$ is symmetric about $0$,$I_1 = 0$.Let $I_2 = \int_{0}^{1} \frac{dt}{t^2 + 2t \cos \alpha + 1}$. We can write the denominator as $(t + \cos \alpha)^2 + \sin^2 \alpha$.Thus,$I_2 = \int_{0}^{1} \frac{dt}{(t + \cos \alpha)^2 + \sin^2 \alpha} = \left[ \frac{1}{\sin \alpha} 	an^{-1} \left( \frac{t + \cos \alpha}{\sin \alpha} ight) ight]_0^1$.$I_2 = \frac{1}{\sin \alpha} \left( 	an^{-1} \left( \frac{1 + \cos \alpha}{\sin \alpha} ight) - 	an^{-1} \left( \frac{\cos \alpha}{\sin \alpha} ight) ight)$.Using half-angle identities,$\frac{1 + \cos \alpha}{\sin \alpha} = \frac{2 \cos^2(\alpha/2)}{2 \sin(\alpha/2) \cos(\alpha/2)} = \cot(\alpha/2) = 	an(\pi/2 - \alpha/2)$.Also,$\frac{\cos \alpha}{\sin \alpha} = \cot \alpha = 	an(\pi/2 - \alpha)$.So,$I_2 = \frac{1}{\sin \alpha} (\pi/2 - \alpha/2 - (\pi/2 - \alpha)) = \frac{1}{\sin \alpha} (\alpha/2) = \frac{\alpha}{2 \sin \alpha}$.The equation becomes $\left( \frac{\alpha}{2 \sin \alpha} ight) x^2 - 0 \cdot x - 2 = 0$.$x^2 = \frac{4 \sin \alpha}{\alpha} \Rightarrow x = \pm 2 \sqrt{\frac{\sin \alpha}{\alpha}}$.

If $x$ satisfies the equation $\left( \int_{0}^{1} \frac{dt}{t^2 + 2t \cos \alpha + 1} \right) x^2 - \left( \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt \right) x - 2 = 0$ for $0 < \alpha < \pi$,then the value of $x$ is

Similar Questions

Which of the following inequalities is/are $TRUE$?
$(A)$ $\int_0^1 x \cos x \, dx \geq \frac{3}{8}$
$(B)$ $\int_0^1 x \sin x \, dx \geq \frac{3}{10}$
$(C)$ $\int_0^1 x^2 \cos x \, dx \geq \frac{1}{2}$
$(D)$ $\int_0^1 x^2 \sin x \, dx \geq \frac{2}{9}$

If $f(x)$ is a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ is a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral $\int_0^1 f(x) g(x) d x$ is

Let $I_n = \int_{0}^{\frac{\pi}{4}} \tan^n x \, dx$. Then $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}, \dots$ are in:

Let $f: R \rightarrow R$ be defined as $f(x)=a e^{2 x}+b e^x+c x$. If $f(0)=-1$,$f^{\prime}(\log _e 2)=21$,and $\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$,then the value of $|a+b+c|$ equals:

If $A_n = \int_{0}^{\pi /2} \frac{\sin((2n-1)x)}{\sin x} dx$ and $B_n = \int_{0}^{\pi /2} \left( \frac{\sin(nx)}{\sin x} \right)^2 dx$ for $n \in N$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module