int_0^{pi / 4} tan ^2(x) , dx = | vedclass

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

Question

(C) We need to evaluate the integral $I = \int_0^{\pi / 4} 	an^2(x) \, dx$. Using the trigonometric identity $	an^2(x) = \sec^2(x) - 1$,we can rewri

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We need to evaluate the integral $I = \int_0^{\pi / 4} 	an^2(x) \, dx$. Using the trigonometric identity $	an^2(x) = \sec^2(x) - 1$,we can rewrite the integral as: $I = \int_0^{\pi / 4} (\sec^2(x) - 1) \, dx$. Now,integrate each term separately: $I = [	an(x) - x]_0^{\pi / 4}$. Applying the limits: $I = (	an(\frac{\pi}{4}) - \frac{\pi}{4}) - (	an(0) - 0)$. Since $	an(\frac{\pi}{4}) = 1$ and $	an(0) = 0$,we get: $I = (1 - \frac{\pi}{4}) - (0 - 0) = 1 - \frac{\pi}{4}$. Thus,the correct option is $C$.

$\int_0^{\pi / 4} \tan ^2(x) \, dx =$

Similar Questions

$\int_0^3 \left(\sin \left(\frac{\pi}{3} x\right) - \cos \left(\frac{\pi}{3} x\right)\right) dx =$

$\int_{-2}^2 |[x]| \, dx$ is equal to

$\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$ is equal to :

$\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=$

If $\int_0^{k} \frac{d x}{2+8 x^2}=\frac{\pi}{16}$,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module