If f(x) = int left[ tan^2 x + cot^2 x + frac{ | vedclass

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

Question

(A) Given $f(x) = \int \left[ 	an^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{(2 \sin x \cos x)^2} ight] dx$. Since $\sin^2 2x = 4 \sin^2 x \cos

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given $f(x) = \int \left[ 	an^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{(2 \sin x \cos x)^2} ight] dx$. Since $\sin^2 2x = 4 \sin^2 x \cos^2 x$, the expression becomes $\int \left[ 	an^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{4 \sin^2 x \cos^2 x} ight] dx$. $= \int \left[ 	an^2 x + \cot^2 x + \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} ight] dx$. $= \int \left[ (\sec^2 x - 1) + (\csc^2 x - 1) + \sec x 	an x + \csc x \cot x ight] dx$. $= 	an x - x - \cot x - x + \sec x - \csc x + C$. $f(x) = 	an x - \cot x + \sec x - \csc x - 2x + C$. Given $f\left(\frac{\pi}{4}ight) = 0$: $	an\frac{\pi}{4} - \cot\frac{\pi}{4} + \sec\frac{\pi}{4} - \csc\frac{\pi}{4} - 2\left(\frac{\pi}{4}ight) + C = 0$. $1 - 1 + \sqrt{2} - \sqrt{2} - \frac{\pi}{2} + C = 0 \implies C = \frac{\pi}{2}$. So $f(x) = 	an x - \cot x + \sec x - \csc x - 2x + \frac{\pi}{2}$. Now $f\left(\frac{\pi}{6}ight) = 	an\frac{\pi}{6} - \cot\frac{\pi}{6} + \sec\frac{\pi}{6} - \csc\frac{\pi}{6} - 2\left(\frac{\pi}{6}ight) + \frac{\pi}{2}$. $= \frac{1}{\sqrt{3}} - \sqrt{3} + \frac{2}{\sqrt{3}} - 2 - \frac{\pi}{3} + \frac{\pi}{2} = \frac{3}{\sqrt{3}} - \sqrt{3} - 2 + \frac{\pi}{6} = \sqrt{3} - \sqrt{3} - 2 + \frac{\pi}{6} = \frac{\pi}{6} - 2$. Then $3 \left[ f\left(\frac{\pi}{6}ight) + 2 ight] = 3 \left[ \frac{\pi}{6} - 2 + 2 ight] = 3 \left( \frac{\pi}{6} ight) = \frac{\pi}{2}$.

If $f(x) = \int \left[ \tan^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{\sin^2 2x} \right] dx$ and $f\left(\frac{\pi}{4}\right) = 0$, then $3 \left[ f\left(\frac{\pi}{6}\right) + 2 \right] = $

Similar Questions

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

If $\int(x+2) \sqrt{x^2-x+2} \, dx = \frac{1}{3} f(x) + \frac{5}{8} g(x) + \frac{35}{16} h(x) + c$, then $f(-1) + g(-1) + h\left(\frac{1}{2}\right) = $

$\int {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx = $

If $\int \frac{1}{(x-2)^5(x-1)^4} d x=\sum_{r=-4}^{-1} A_r\left(\frac{x-2}{x-1}\right)^r+\sum_{r=1}^3 A_r\left(\frac{x-2}{x-1}\right)^r+B f(x)$, then $f(x)=$

If $\int \frac{1+x^2}{1+x^4} dx=\frac{1}{\sqrt{2}} \tan ^{-1}\left[\frac{f(x)}{\sqrt{2}}\right]+c$,then $f(x)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module