The value of the int_{0}^{frac{pi}{2}} left( | vedclass

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

Question

(D) We know that $\sin 3y = 3\sin y - 4\sin^3 y$. Substituting this into the integrand: $\frac{1 + 3\sin y - 4\sin^3 y}{1 + 2\sin y}$. Using polynomia

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We know that $\sin 3y = 3\sin y - 4\sin^3 y$. Substituting this into the integrand: $\frac{1 + 3\sin y - 4\sin^3 y}{1 + 2\sin y}$. Using polynomial division or factoring: $1 + 3\sin y - 4\sin^3 y = (1 + 2\sin y)(1 + \sin y - 2\sin^2 y)$. Thus,the integrand simplifies to $1 + \sin y - 2\sin^2 y$. Now,integrate with respect to $y$ from $0$ to $\frac{\pi}{2}$: $\int_{0}^{\frac{\pi}{2}} (1 + \sin y - 2\sin^2 y) dy = \int_{0}^{\frac{\pi}{2}} (1 + \sin y - (1 - \cos 2y)) dy$. $= \int_{0}^{\frac{\pi}{2}} (\sin y + \cos 2y) dy$. $= [-\cos y + \frac{1}{2}\sin 2y]_{0}^{\frac{\pi}{2}}$. $= (-\cos \frac{\pi}{2} + \frac{1}{2}\sin \pi) - (-\cos 0 + \frac{1}{2}\sin 0)$. $= (0 + 0) - (-1 + 0) = 1$.

The value of the $\int_{0}^{\frac{\pi}{2}} \left( \frac{1 + \sin 3y}{1 + 2\sin y} \right) dy$ is equal to

Similar Questions

$\int_0^2 \sqrt{\frac{2 + x}{2 - x}} \,dx = $

Let $[t]$ denote the greatest integer function. If $\int_0^{2.4} [x^2] dx = \alpha + \beta \sqrt{2} + \gamma \sqrt{3} + \delta \sqrt{5}$,then $\alpha + \beta + \gamma + \delta$ is equal to $..............$.

If $I_n = \int_0^{\pi/4} \tan^n \theta \, d\theta$ for $n = 1, 2, 3, \ldots$,then $I_{n-1} + I_{n+1}$ is equal to

Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$.

$\int_0^\infty {\frac{{x\,dx}}{{(1 + x)(1 + {x^2})}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module