int frac{sin^3(x) + cos^3(x)}{sin^2(x) cdot c | vedclass

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

Question

(A) We are given the integral: $I = \int \frac{\sin^3(x) + \cos^3(x)}{\sin^2(x) \cos^2(x)} \, dx$ Divide each term in the numerator by the denominator

Vedclass · Accepted Answer

$\sec(x) - \operatorname{cosec}(x) + C$

Vedclass · Answer

(A) We are given the integral: $I = \int \frac{\sin^3(x) + \cos^3(x)}{\sin^2(x) \cos^2(x)} \, dx$ Divide each term in the numerator by the denominator: $I = \int \left( \frac{\sin^3(x)}{\sin^2(x) \cos^2(x)} + \frac{\cos^3(x)}{\sin^2(x) \cos^2(x)} ight) \, dx$ $I = \int \left( \frac{\sin(x)}{\cos^2(x)} + \frac{\cos(x)}{\sin^2(x)} ight) \, dx$ Using trigonometric identities $\frac{\sin(x)}{\cos(x)} = 	an(x)$ and $\frac{1}{\cos(x)} = \sec(x)$,and $\frac{\cos(x)}{\sin(x)} = \cot(x)$ and $\frac{1}{\sin(x)} = \operatorname{cosec}(x)$: $I = \int (	an(x) \sec(x) + \cot(x) \operatorname{cosec}(x)) \, dx$ Integrating term by term: $\int \sec(x) 	an(x) \, dx = \sec(x)$ $\int \operatorname{cosec}(x) \cot(x) \, dx = -\operatorname{cosec}(x)$ Thus,$I = \sec(x) - \operatorname{cosec}(x) + C$. Hence,option $A$ is correct.

$\int \frac{\sin^3(x) + \cos^3(x)}{\sin^2(x) \cdot \cos^2(x)} \, dx = $

Similar Questions

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

Find an anti-derivative (or integral) of the function $\cos 3x$ by the method of inspection.

If $\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx = A \sin 2 x + B \sin 4 x + C \sin 6 x + D \sin 8 x + k$ (where $k$ is the arbitrary constant of integration),then $\frac{1}{B} + \frac{1}{C} = $

$\int \frac{x^3+2 x}{x^4+4} d x=$

Find: $\int \sin^{3} x \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module