int_0^{pi} (sin^2 frac{x}{2} - cos^2 frac{x}{ | vedclass

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

Question

(B) We know that the trigonometric identity is $\cos^2 	heta - \sin^2 	heta = \cos 2	heta$.Therefore,$\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) We know that the trigonometric identity is $\cos^2 	heta - \sin^2 	heta = \cos 2	heta$.Therefore,$\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos(2 \cdot \frac{x}{2}) = -\cos x$.Now,the integral becomes $I = \int_0^{\pi} -\cos x \, dx$.Integrating $-\cos x$ with respect to $x$,we get $-\sin x$.Applying the limits from $0$ to $\pi$,we have $I = [-\sin x]_0^{\pi}$.$I = -(\sin \pi - \sin 0)$.Since $\sin \pi = 0$ and $\sin 0 = 0$,we get $I = -(0 - 0) = 0$.

$\int_0^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) dx = $ . . . . . . .

Similar Questions

The value of the definite integral,$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin 5x}}{{\sin x}}\,dx} $ is

Let $(2^{1-a} + 2^{1+a})$,$f(a)$,$(3^a + 3^{-a})$ be in $A$.$P$. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$ is:

$\int_0^{\pi / 4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ is equal to

Evaluate the definite integral $\int_{1}^{2} (4x^{3} - 5x^{2} + 6x + 9) dx$.

If for all real triplets $(a, b, c)$,$f(x) = a + bx + cx^2$,then $\int_{0}^{1} f(x) dx$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module