If int sec ^4 x cdot tan ^4 x , dx = frac{tan | vedclass

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

Question

(B) Let $I = \int \sec ^4 x 	an ^4 x \, dx$. We can rewrite the integral as $I = \int \sec ^2 x \cdot \sec ^2 x \cdot 	an ^4 x \, dx$. Using the ide

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Let $I = \int \sec ^4 x 	an ^4 x \, dx$. We can rewrite the integral as $I = \int \sec ^2 x \cdot \sec ^2 x \cdot 	an ^4 x \, dx$. Using the identity $\sec ^2 x = 1 + 	an ^2 x$,we get $I = \int (1 + 	an ^2 x) 	an ^4 x \cdot \sec ^2 x \, dx$. Let $	an x = t$,then $\sec ^2 x \, dx = dt$. Substituting these into the integral,we get $I = \int (1 + t^2) t^4 \, dt = \int (t^4 + t^6) \, dt$. Integrating with respect to $t$,we obtain $I = \frac{t^5}{5} + \frac{t^7}{7} + c$. Substituting back $t = 	an x$,we have $I = \frac{	an ^5 x}{5} + \frac{	an ^7 x}{7} + c$. Comparing this with the given form $\frac{	an ^m x}{m} + \frac{	an ^n x}{n} + c$,we get $m = 5$ and $n = 7$ (or vice versa). Therefore,$m + n = 5 + 7 = 12$.

If $\int \sec ^4 x \cdot \tan ^4 x \, dx = \frac{\tan ^m x}{m} + \frac{\tan ^n x}{n} + c$ (where $c$ is the constant of integration),then $m + n =$

Similar Questions

$\int \frac{d x}{(x+1) \sqrt{4 x+3}}$ is equal to

Consider $f(x) = \frac{x^2}{1 + x^3}$ and $g(t) = \int f(t) \, dt$. If $g(1) = 0$,then $g(x)$ equals:

$\int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} d x$ is equal to

$\int \frac{e^{\tan ^{-1} 2 x}}{1+4 x^2} dx =$

If $\int \frac{dx}{1+3 \sin^2 x} = \frac{1}{2} \tan^{-1}(f(x)) + c$,where $c$ is a constant of integration,then $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module