Question

If $\int \frac{2x^{2}se{c}^{2}xdx}{(xse{c}^{2}x-tanx{)}^2}=f\left(x\right)+cosx+x+c$, where C is constant of integration, then value of $f\left(\frac{\pi }{4}\right)-f(-\frac{\pi }{4})$ is equal to

• A. $\frac{\pi }{2+\pi }$
• B. $\frac{\pi }{2-\pi }$
• C. $\frac{\pi }{4-\pi }$
• D. $\frac{\pi }{4+\pi }$

Answer 1

The correct option is B

$\frac{\pi }{2-\pi }$

$\int \underset{ii}{\frac{2x^{2}se{c}^{2}xdx}{(xse{c}^{2}x-tanx{)}^2}}\underset{i}{\frac{x}{tanx}}dx$

$=\frac{x}{tanx}.\frac{1}{(xse{c}^{2}x-tanx)}-\int \frac{tanx-se{c}^2}{ta{n}^{2}x}\frac{-dx}{xse{c}^{2}x-tanx}=\frac{-x}{\left(tanx\right)(xse{c}^{2}x-tanx)}-\int co{t}^{2}xdx=\frac{-x}{\left(tanx\right)(xse{c}^{2}x-tanx)}+cotx+x+c\Rightarrow f\left(x\right)=\frac{-x}{\left(tanx\right)(xse{c}^{2}x-tanx)}$