If f(x) = int frac{x^2 + sin^2 x}{1 + x^2} cd | vedclass

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

Question

(D) Given $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$. We can rewrite the numerator as $x^2 + 1 - 1 + \sin^2 x = (1 + x^2) - (1

Vedclass · Accepted Answer

$	an 1 - \frac{\pi}{4}$

Vedclass · Answer

(D) Given $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$. We can rewrite the numerator as $x^2 + 1 - 1 + \sin^2 x = (1 + x^2) - (1 - \sin^2 x) = (1 + x^2) - \cos^2 x$. Substituting this into the integral: $f(x) = \int \frac{(1 + x^2) - \cos^2 x}{1 + x^2} \cdot \sec^2 x \, dx$ $f(x) = \int \left( 1 - \frac{\cos^2 x}{1 + x^2} ight) \sec^2 x \, dx$ $f(x) = \int \sec^2 x \, dx - \int \frac{\cos^2 x \cdot \sec^2 x}{1 + x^2} \, dx$ Since $\cos^2 x \cdot \sec^2 x = 1$,we have: $f(x) = \int \sec^2 x \, dx - \int \frac{1}{1 + x^2} \, dx$ $f(x) = 	an x - 	an^{-1} x + C$. Given $f(0) = 0$: $0 = 	an(0) - 	an^{-1}(0) + C \implies 0 = 0 - 0 + C \implies C = 0$. Thus,$f(x) = 	an x - 	an^{-1} x$. Therefore,$f(1) = 	an(1) - 	an^{-1}(1) = 	an(1) - \frac{\pi}{4}$.

If $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$ and $f(0) = 0$,then $f(1) = $

Similar Questions

If $\int \frac{1}{\left((x+4)^3(x+1)^5\right)^{1 / 4}} d x=A \cdot\left(\frac{x+4}{x+1}\right)^n+c$,then which of the following is true?

The value of $\int e^{\tan ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$ is

$\int \operatorname{cosec}(x-a) \operatorname{cosec} x \, dx =$

Integrate the function $\frac{1}{1+\cot x}$.

The value of $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module