int frac{sin^2 x - cos^2 x}{sin^2 x cos^2 x} | vedclass

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

Question

(A) Given integral: $I = \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ Split the fraction: $I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x

Vedclass · Accepted Answer

$	an x + \cot x + c$

Vedclass · Answer

(A) Given integral: $I = \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ Split the fraction: $I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x} ight) dx$ Simplify the terms: $I = \int \left( \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} ight) dx$ Use trigonometric identities: $I = \int (\sec^2 x - \csc^2 x) dx$ Integrate: $I = 	an x - (-\cot x) + c$ Final result: $I = 	an x + \cot x + c$

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

Similar Questions

The value of $\int {\frac{1}{{{{(x - 5)}^2}}}\,dx} $ is

Find the following integral:
$\int \frac{1-\sin x}{\cos ^{2} x} d x$

If $f(\frac{x - 1}{x + 1}) = x + 1$, then $\int f(x) dx = \dots$

$\int \frac{x^4+x^2+1}{x^2-x+1} \,d x$ is equal to

Find the integral of the function $\sin ^{2}(2 x+5)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module