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

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

Question

(A) Given $f(x) = \int \left( \frac{x^2 + \sin^2 x}{1 + x^2} \right) \sec^2 x \, dx$.We can rewrite the integrand as:$f(x) = \int \frac{x^2 \sec^2 x +

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $f(x) = \int \left( \frac{x^2 + \sin^2 x}{1 + x^2} ight) \sec^2 x \, dx$.We can rewrite the integrand as:$f(x) = \int \frac{x^2 \sec^2 x + \sin^2 x \sec^2 x}{1 + x^2} \, dx$Since $\sin^2 x \sec^2 x = 	an^2 x$,we have:$f(x) = \int \frac{x^2 \sec^2 x + 	an^2 x}{1 + x^2} \, dx$Using $\sec^2 x = 1 + 	an^2 x$,we get:$f(x) = \int \frac{x^2(1 + 	an^2 x) + 	an^2 x}{1 + x^2} \, dx$$f(x) = \int \frac{x^2 + x^2 	an^2 x + 	an^2 x}{1 + x^2} \, dx = \int \frac{x^2 + 	an^2 x(1 + x^2)}{1 + x^2} \, dx$$f(x) = \int \frac{x^2}{1 + x^2} \, dx + \int 	an^2 x \, dx$$f(x) = \int \frac{x^2 + 1 - 1}{1 + x^2} \, dx + \int (\sec^2 x - 1) \, dx$$f(x) = \int (1 - \frac{1}{1 + x^2}) \, dx + \int \sec^2 x \, dx - \int 1 \, dx$$f(x) = x - 	an^{-1} x + 	an x - x + C = 	an x - 	an^{-1} x + C$Given $f(0) = 0$,we have $0 = 	an 0 - 	an^{-1} 0 + C$,so $C = 0$.Thus,$f(x) = 	an x - 	an^{-1} x$.Therefore,$f(1) = 	an 1 - 	an^{-1}(1) = 	an 1 - \frac{\pi}{4}$.

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

Similar Questions

If $\int {({x^3} - 2{x^2} + 5){e^{3x}}\,dx} = e^{3x} (Ax^3 + Bx^2 + Cx + D) + K$,then the statement which is incorrect is

Find $\int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx$

If $\int \frac{1}{\cos 4x \cos 2x} dx = \frac{1}{2\sqrt{2}} \log \left(\frac{1+f(x)}{1-f(x)}\right) - \frac{1}{2} \log g(x) + C$,then find the value of $g\left(\frac{\pi}{6}\right) - \sqrt{2} f\left(\frac{\pi}{6}\right)$.

$\int \left[ \log (1+\cos x) - x \tan \left( \frac{x}{2} \right) \right] dx =$

If $\int \frac{d x}{\cot ^2 x-1}=\frac{1}{A} \log |\sec 2 x+\tan 2 x|-\frac{x}{B}+c$,(where $c$ is the constant of integration),then $A+B=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module