\mathbf{(1)\;\;\int\frac{1}{\left(\sec x+\csc x+\tan x+\cot x\right)^2}dx}
\mathbf{(2)\;\int_{0}^{\pi}\left\{\mid \sin (2010x)\mid}-\mid \sin (2011x)\mid\right\}dx$
1for the first one you will have to convert all sec x .... into sin and cos form doing so you wl gt a result like this ......
∫sinx cosx
1+sinx+cosx
1/2∫2sinx cosx
1+sinx+cosx
convert the sinx+cos x in denominator in the form of √2sin(Λ/4+x)
and numerator bcoms sin2x
hence sustitute ^/4+x as z and the numerator bcoms -cos (2z) where as in denominator we have 1+sinz now i think you could integrate it further if you still dnt get it then post me
1708\hspace{-16}(1)::\int\frac{1}{\left(\sec x+\csc x+\tan x+\cot x\right)^2}dx\\\\\\ $simplify $\frac{1}{\left(\sec x+\csc x+\tan x+\cot x\right)^2} = \frac{1}{\left(\frac{1}{\cos x}+\frac{1}{\sin x}+\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)^2}$\\\\\\ $= \frac{\sin^2 x.\cos^2 x}{2.\left(1+\sin x+\cos x\right) } = \frac{1}{2}.\left(1-\cos x-\sin x+\sin x.\cos x\right)$\\\\\\ because $\sin^2 x.\cos^2 x = \left(1-\cos^2 x\right).\left(1-\sin^2 x\right) = (1+\sin x).(1+\cos x).(1-\sin x).(1-\cos x)$\\\\\\ $= \left(1+\cos x+\sin x+\sin x.\cos x\right).\left(1-\cos x-\sin x+\sin x.\cos x\right)$\\\\\\ so $\int\frac{1}{\left(\sec x+\csc x+\tan x+\cot x\right)^2}\,dx =\frac{1}{2}.\int \left(1-\cos x-\sin x+\sin x.\cos x\right)\,dx$\\\\\\ $= \frac{1}{2}.\left(x-\sin x+\cos x\right)-\frac{1}{8}.\cos 2x +C$
For (2)::::::
