If int f(x) sin x cos x , dx = frac{1}{2(b^2 | vedclass

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

Question

(C) Given $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log(f(x)) + c$. Differentiating both sides with respect to $x$,we get: $f(x) \sin x

Vedclass · Accepted Answer

$\frac{1}{a^2 \cos^2 x + b^2 \sin^2 x}$

Vedclass · Answer

(C) Given $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log(f(x)) + c$. Differentiating both sides with respect to $x$,we get: $f(x) \sin x \cos x = \frac{1}{2(b^2 - a^2)} \cdot \frac{f'(x)}{f(x)}$. Rearranging the terms: $2(b^2 - a^2) \sin x \cos x = \frac{f'(x)}{f(x)^2}$. Integrating both sides with respect to $x$: $\int (2b^2 \sin x \cos x - 2a^2 \sin x \cos x) \, dx = \int \frac{f'(x)}{f(x)^2} \, dx$. Using the substitution $u = \sin^2 x$ and $v = \cos^2 x$,we have $du = 2 \sin x \cos x \, dx$ and $dv = -2 \cos x \sin x \, dx$. Thus,$b^2 \sin^2 x - a^2 \cos^2 x = -\frac{1}{f(x)} + C_1$. Simplifying,we get $f(x) = \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x}$ (adjusting constants). Comparing with the given options,the correct form is $f(x) = \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x}$.

If $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log(f(x)) + c$,then $f(x) = $

Similar Questions

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

$\int \frac{x+\cos x}{1-\sin x} d x=$

$\int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} dx =$

$\int \frac{1}{\sqrt{1 + \sin x}} \, dx = $

$\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module