Evaluation of $\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}dx$ 
Evaluation of $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}dx$$

$\bf{My\; Try::}$ We can write $$x\cos x+\sin x= \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \cos x+\frac{1}{\sqrt{1+x^2}}\cdot \sin x\right\}$$
So we get $$(x\cos x+\sin x) = \sqrt{1+x^2}\cos(x-\alpha)\;,$$ Where $\displaystyle \alpha = \tan^{-1}\left(\frac{1}{x}\right)$
So Integral $$I = \int\frac{(1+x^2)(2+x^2)}{(1+x^2)^2\cdot \cos^4 (x-\alpha)}dx = \int\frac{(2+x^2)}{(1+x^2)}\cdot \sec^4 (x-\alpha)dx$$
Now how can i solve after that,Help me
Thanks
 A: Your integral becomes $$I=\int { \frac { \left( 2+{ x }^{ 2 } \right)  }{ \left( 1+{ x }^{ 2 } \right)  } { \left( \sec { \left( x-\cot ^{ -1 }{ x }  \right)  }  \right)  }^{ 4 } } dx$$
On substituting $$\left( x-\cot ^{ -1 }{ x }  \right) =t$$
diffrentiating $$dt = \frac { \left( 2+{ x }^{ 2 } \right)  }{ \left( 1+{ x }^{ 2 } \right)  } dx$$
Integration becomes $$I=\int { { \sec { \left( t \right)  }  }^{ 4 } } dt$$

Using :      $\quad \quad \sec ^{ 2 }{ \left( x \right)  } -\tan ^{ 2 }{ \left( x \right)  } =1$

$$I=\int { { \sec { \left( t \right)  }  }^{ 2 } } \left( { \tan { \left( t \right)  }  }^{ 2 }+1 \right) dt$$
Again substituting $$\tan{ \left (t\right)} = u$$
$$du = { { \sec { \left( t \right)  }  }^{ 2 } }dt$$
Intrgration becomes $$I = \int { \left( { u }^{ 2 }+1 \right) du } $$
$$I=\frac { { u }^{ 3 } }{ 3 } +u+c$$
Substituting value of u 
$$I = \frac { { \tan { \left( t \right)  }  }^{ 3 } }{ 3 } +\tan { \left( t \right)  } +c$$
$$I=\frac { { \tan { \left( x-\cot ^{ -1 }{ x }  \right)  }  }^{ 3 } }{ 3 } +\tan { \left( x-\cot ^{ -1 }{ x }  \right)  } +c$$
