int_0^{frac{pi}{4}} sec^4 x , dx = | vedclass

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

Question

(C) We want to evaluate the integral $I = \int_0^{\frac{\pi}{4}} \sec^4 x \, dx$. Using the trigonometric identity $\sec^2 x = 1 + 	an^2 x$,we can re

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(C) We want to evaluate the integral $I = \int_0^{\frac{\pi}{4}} \sec^4 x \, dx$. Using the trigonometric identity $\sec^2 x = 1 + 	an^2 x$,we can rewrite the integral as: $I = \int_0^{\frac{\pi}{4}} \sec^2 x \cdot \sec^2 x \, dx = \int_0^{\frac{\pi}{4}} (1 + 	an^2 x) \sec^2 x \, dx$. Let $u = 	an x$. Then $du = \sec^2 x \, dx$. When $x = 0$,$u = 	an(0) = 0$. When $x = \frac{\pi}{4}$,$u = 	an(\frac{\pi}{4}) = 1$. Substituting these into the integral: $I = \int_0^1 (1 + u^2) \, du$. Integrating with respect to $u$: $I = [u + \frac{u^3}{3}]_0^1$. Evaluating at the limits: $I = (1 + \frac{1^3}{3}) - (0 + \frac{0^3}{3}) = 1 + \frac{1}{3} = \frac{4}{3}$.

$\int_0^{\frac{\pi}{4}} \sec^4 x \, dx =$

Similar Questions

If $f(x) = \begin{cases} \frac{6 x^2 + 1}{4 x^3 + 2 x + 3}, & 0 < x < 1 \\ x^2 + 1, & 1 \le x \le 2 \end{cases}$ then $\int_0^2 f(x) dx =$

$\int_0^1 \tan^{-1} x \, dx =$

The value of the definite integral,$\int\limits_0^{100} {\frac{x}{{{e^{{x^2}}}}}} \,dx$ is equal to

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

$\int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module