Let F(x) be an indefinite integral of sin ^2 | vedclass

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

Question

(D) Given $F(x) = \int \sin^2 x \, dx$.Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we have:$F(x) = \int \frac{1 - \cos 2x}{2} \, dx = \frac{

Vedclass · Accepted Answer

Statement -$1$ is False,Statement -$2$ is True

Vedclass · Answer

(D) Given $F(x) = \int \sin^2 x \, dx$.Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we have:$F(x) = \int \frac{1 - \cos 2x}{2} \, dx = \frac{1}{2} x - \frac{1}{4} \sin 2x + C$.Now,check $F(x + \pi)$:$F(x + \pi) = \frac{1}{2}(x + \pi) - \frac{1}{4} \sin(2(x + \pi)) + C$$F(x + \pi) = \frac{1}{2}x + \frac{\pi}{2} - \frac{1}{4} \sin(2x + 2\pi) + C$$F(x + \pi) = \frac{1}{2}x + \frac{\pi}{2} - \frac{1}{4} \sin 2x + C = F(x) + \frac{\pi}{2}$.Since $F(x + \pi) 
eq F(x)$,Statement-$1$ is False.For Statement-$2$,we know $\sin(x + \pi) = -\sin x$,so $\sin^2(x + \pi) = (-\sin x)^2 = \sin^2 x$. Thus,Statement-$2$ is True.Therefore,Statement-$1$ is False and Statement-$2$ is True.

Let $F(x)$ be an indefinite integral of $\sin ^2 x$.
$STATEMENT -1$ : The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$. because
$STATEMENT -2$: $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.

Similar Questions

Let $g:(0, \infty) \rightarrow R$ be a differentiable function such that $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} \right) dx = \frac{x g(x)}{e^x + 1} + c$ for all $x > 0$,where $c$ is an arbitrary constant. Then:

$\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x(1+\cos ^2 2 x)} d x=$

$\int \frac{\tan^{-1} x - \cot^{-1} x}{\tan^{-1} x + \cot^{-1} x} \, dx$ is equal to:

If $I_n = \int \tan^n x \ dx$,and $I_0 + I_1 + 2 I_2 + 2 I_3 + 2 I_4 + I_5 + I_6 = \sum_{K=1}^n \frac{\tan^K x}{K}$,then $n = $

If $\int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx = \frac{1}{12} \tan^{-1}(3 \tan x) + C$,then the maximum value of $a \sin x + b \cos x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $F(x)$ be an indefinite integral of $\sin ^2 x$.$STATEMENT -1$ : The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$. because$STATEMENT -2$: $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.