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

$\int \frac{1}{1+x+x^2} \, dx =$

$\int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x=$

$\int \tan ^{-1}(\sec x+\tan x) d x=$

$\int \frac{1}{9 \cos ^2 x-24 \sin x \cos x+16 \sin ^2 x} d x=$

Find the following integral: $\int \frac{dx}{\sqrt{5x^{2}-2x}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module