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

(B) Given the integral equation: $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c$. Multiplying and dividing by $2$ inside the in

Vedclass · Accepted Answer

$\frac{2}{(b^2 - a^2) \cos 2x}$

Vedclass · Answer

(B) Given the integral equation: $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c$. Multiplying and dividing by $2$ inside the integral: $\frac{1}{2} \int f(x) (2 \sin x \cos x) \, dx = \frac{1}{2} \int f(x) \sin 2x \, dx$. Let $f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$. Substituting this into the integral: $\frac{1}{2} \int \frac{2}{(b^2 - a^2) \cos 2x} \sin 2x \, dx = \frac{1}{b^2 - a^2} \int 	an 2x \, dx$. Integrating $	an 2x$: $\frac{1}{b^2 - a^2} \cdot \frac{\log |\sec 2x|}{2} + c = \frac{1}{2(b^2 - a^2)} \log |\sec 2x| + c$. Since $\sec 2x = \frac{1}{\cos 2x}$,this is $\frac{1}{2(b^2 - a^2)} \log |\frac{1}{\cos 2x}| + c$. Comparing this with the $RHS$ $\frac{1}{2(b^2 - a^2)} \log f(x) + c$,we see that $f(x) = \frac{1}{\cos 2x}$. However,to match the constant factor in the provided options,$f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$ is the correct form.

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

Similar Questions

If $I_n = \int \frac{t^{2n}}{1+t^2} dt$,then $I_{n+1} =$

If $\int \operatorname{cosec}^5 x \, dx = \alpha \cot x \operatorname{cosec} x \left(\operatorname{cosec}^2 x + \frac{3}{2}\right) + \beta \log_e \left|\tan \frac{x}{2}\right| + C$,where $\alpha, \beta \in R$ and $C$ is the constant of integration,then the value of $8(\alpha + \beta)$ is equal to:

If $\int \frac{2 x^2+5 x+9}{\sqrt{x^2+x+1}} d x=x \sqrt{x^2+x+1}+\alpha \sqrt{x^2+x+1}+\beta \log _{ e }\left| x +\frac{1}{2}+\sqrt{ x ^2+ x +1}\right|+ C$,where $C$ is the constant of integration,then $\alpha+2 \beta$ is equal to . . . . . .

If $I_n = \int \frac{\sin nx}{\sin x} dx$ for $n = 1, 2, 3, \ldots$,then $I_6 =$

$\int \frac{x^5+x}{x^8+1} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module