If ∫$$cos4x+1cotx-tanx$$ $$dx=Acos4x+B$$, then the value of A is

∫$$2$$ $$cos22x2$$ $$cot$$ $$2x=$$∫ $$cos22x$$ $$sin2xcos2xdx$$ $$=12$$∫$$sin4x$$ dx $$=-18cos4x+B$$

Methods of integration

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Question

$If ∫ \frac{c o s 4 x + 1}{c o t x - t a n x} d x = A c o s 4 x + B, t h e n t h e v a l u e o f A i s$

Easy

Solution

$\int \frac{2 \cos^{2} 2 x}{2 c o t 2 x} = \int_{}^{} \frac{c o s^{2} 2 x s i n 2 x}{c o s 2 x} d x = \frac{1}{2} \int s i n 4 x d x = \frac{- 1}{8} c o s 4 x + B$

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