Discussion Forum

Let   $f(x)= 7\tan^{8}x+7\tan^{6}x-3\tan^{4}x$ $x-3\tan^{2}x$   for all   $x\epsilon (-\frac{\pi}{2},\frac{\pi}{2})$  , Then the correct expression (s) is /are

Answer:

Explanation:

Here,      $f(x)= 7\tan^{8}x+7\tan^{6}x-3\tan^{4}x-3\tan^{2}x$

   For all x   $(\frac{-\pi}{2},\frac{\pi}{2})$

 $\therefore$          $f(x)= 7\tan^{6}x \sec^{2}x-3\tan^{2}x \sec^{2}x$

= $(7\tan^{6}x-3\tan^{2}x)\sec^{2}x$

 Now,     $\int_{0}^{\frac{\pi}{4}} x f(x) dx=\int_{0}^{\frac{\pi}{4}} x(7\tan^{6}x-3\tan^{2}x)\sec^{2}x dx$

                      = $\left[ x(\tan^{7}x-\tan^{3}x)\right]_0^{\pi/4}$

                                                         - $\int_{0}^{\pi/4} 1(tan^{7}x-\tan^{3}x)dx$

  = 0-  $\int_{0}^{\pi/4} tan^{3}x(tan^{4}x-1)dx$

     =   $-\int_{0}^{\pi/4} tan^{3}x(tan^{2}x-1)\sec^{2}xdx$

Put tan x=t   $\Rightarrow \sec^{2}x dx=dt$

    $\therefore$           $\int_{0}^{\pi/4}x f(x)=-\int_{0}^{1} t^{3}(t^{2}-1)dt$

   $=\int_{0}^{1} (t^{b}-t^{5})dt=\left[\frac{t^{4}}{4}-\frac{t^{5}}{5}\right]_0^1=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$

 Also, 

         $\int_{0}^{\pi/4} f(x) dx= \int_{0}^{\pi/4} (7\tan^{6}x-3\tan^{2}x)\sec^{2}x dx$

            =    $\int_{0}^{1}   (7t^{6}-3t^{2}) dt=\left[t^{7}-t^{3}\right]_0^1=0$

                                      -