Suppose ∫1−7cos2⁡xsin7⁡xcos2⁡xdx=g(x)sin7⁡x+C, where C is an arbitrary constant of integration. Then  the value of g′(0)+g''π4.

∫1−7cos2xsin7x⋅cos2xdx=∫cosec7xsec2x−7∫cosec7xdx=cosec7x(tanx)−∫7cosec6x−cosecx(cotx)⋅(tanx)−7∫cosec7xdx=tanxsin7x+∫7cosec7xdx−7∫cosec7xdx=tanxsin7x∴gx=tanxg'(x)=sec2x, g''(x)=2sec2xtanx. g'(0)=1 g''π4=2(2)(1)=4.g'(0)+g''π4=5