int {{e^x}(1 - cot x + {{cot }^2}x),dx} equa | vedclass

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

Question

(C) We know that $1 + \cot^2 x = \csc^2 x$. Substituting this into the integral,we get: $I = \int e^x (1 - \cot x + \cot^2 x) \, dx = \int e^x (\csc^2

Vedclass · Accepted Answer

$-{e^x}\cot x + c$

Vedclass · Answer

(C) We know that $1 + \cot^2 x = \csc^2 x$. Substituting this into the integral,we get: $I = \int e^x (1 - \cot x + \cot^2 x) \, dx = \int e^x (\csc^2 x - \cot x) \, dx$. Let $f(x) = -\cot x$. Then $f'(x) = -(-\csc^2 x) = \csc^2 x$. Using the standard formula $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + c$,we have: $I = e^x(-\cot x) + c = -e^x \cot x + c$.

$\int {{e^x}(1 - \cot x + {{\cot }^2}x)\,dx} $ equals

Similar Questions

$\int e^{-2 x}\left(\tan 2 x-2 \sec ^2 2 x \tan 2 x\right) d x=$

The value of $\int \frac{x e^{x} d x}{(1+x)^{2}}$ is equal to

$\int 2^{x} [f^{\prime}(x) + f(x) \log 2] \, dx$ is equal to

$\int \frac{e^{\sin x}(\sin 2x - 8 \cos x)}{2(\sin x - 3)^2} dx =$

$\int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} \,d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module