If f(x) = int frac{2-3 sin^2 x}{1+cos 2x} dx | vedclass

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

Question

(A) Given $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$. Using the identity $1+\cos 2x = 2 \cos^2 x$,we have: $f(x) = \int \frac{2-3 \sin^2 x}{2 \co

Vedclass · Accepted Answer

$\frac{3}{8}(4-\pi)$

Vedclass · Answer

(A) Given $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$. Using the identity $1+\cos 2x = 2 \cos^2 x$,we have: $f(x) = \int \frac{2-3 \sin^2 x}{2 \cos^2 x} dx = \int \left( \frac{2}{2 \cos^2 x} - \frac{3 \sin^2 x}{2 \cos^2 x} ight) dx$ $f(x) = \int (\sec^2 x - \frac{3}{2} 	an^2 x) dx$ Since $	an^2 x = \sec^2 x - 1$,we get: $f(x) = \int (\sec^2 x - \frac{3}{2}(\sec^2 x - 1)) dx = \int (\sec^2 x - \frac{3}{2} \sec^2 x + \frac{3}{2}) dx$ $f(x) = \int (\frac{3}{2} - \frac{1}{2} \sec^2 x) dx = \frac{3}{2} x - \frac{1}{2} 	an x + C$ Given $f\left(\frac{\pi}{4}ight) = 1$: $\frac{3}{2} \left(\frac{\pi}{4}ight) - \frac{1}{2} 	an\left(\frac{\pi}{4}ight) + C = 1$ $\frac{3\pi}{8} - \frac{1}{2}(1) + C = 1 \Rightarrow C = 1 + \frac{1}{2} - \frac{3\pi}{8} = \frac{3}{2} - \frac{3\pi}{8} = \frac{12-3\pi}{8} = \frac{3}{8}(4-\pi)$ Now,$f(0) = \frac{3}{2}(0) - \frac{1}{2} 	an(0) + C = C = \frac{3}{8}(4-\pi)$.

If $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$ and $f\left(\frac{\pi}{4}\right) = 1$,then $f(0) =$

Similar Questions

$\int \frac{dx}{a^2 - x^2}$ is equal to

$\int \tan(3x - 5) \sec(3x - 5) \, dx = $

Let $f(x) = \frac{1}{x} \ln \left( \frac{x}{e^x} \right)$,then its primitive with respect to $x$ is:

If $\int \sin 5x \cos 3x \; dx = - \frac{\cos 8x}{16} + A$,then $A = $

$\int \frac{\sin 4x}{\sin x} \, dx =$ (where $C$ is a constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module