int (sec^4 x + tan^4 x) , dx = | vedclass

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

Question

(C) Let $I = \int (\sec^4 x + 	an^4 x) \, dx$.
Using the identity $	an^2 x = \sec^2 x - 1$,we have $	an^4 x = (\sec^2 x - 1)^2 = \sec^4 x - 2 \sec

Vedclass · Accepted Answer

$\frac{2}{3} 	an^3 x + 	an x + c$

Vedclass · Answer

(C) Let $I = \int (\sec^4 x + 	an^4 x) \, dx$.
Using the identity $	an^2 x = \sec^2 x - 1$,we have $	an^4 x = (\sec^2 x - 1)^2 = \sec^4 x - 2 \sec^2 x + 1$.
Substituting this into the integral:
$I = \int (\sec^4 x + \sec^4 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^4 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^2 x \cdot \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2(1 + 	an^2 x) \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^2 x + 2 	an^2 x \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 	an^2 x \sec^2 x + 1) \, dx$
Let $u = 	an x$,then $du = \sec^2 x \, dx$.
$I = \int (2u^2 + 1) \, du = \frac{2}{3} u^3 + u + c$
$I = \frac{2}{3} 	an^3 x + 	an x + c$.

$\int (\sec^4 x + \tan^4 x) \, dx = $

Similar Questions

If $\int \frac{x + 1}{x^2 + 1} dx = \tan^{-1} x + g(x) + c$, where $c$ is the constant of integration, then the function $g(x)$ is monotonically increasing in the interval

If $\int \frac{1}{1 + \sin x} dx = \tan(f(x)) + c$,then $f'(0) =$

Evaluate the integral: $\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} dx$ (where $C$ is the constant of integration).

$\int \frac{dx}{\sin^2 x \cos^2 x} = $

If $\int \frac{1}{\sqrt{9-16 x^2}} d x=\alpha \sin ^{-1}(\beta x)+c$,then $\alpha+\frac{1}{\beta}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module