If ( 1-5 cos^2x sin^5x cos^2x ) dx=f (x )+C, where C is the constant of integration, then f ( 6 )-f ( 4 ) is equal to

Question

Accepted Answer

Correct answer: C. (c) 1-5 cos^2x sin^5x cos^2x dx= dx sin^5x cos^2x-5 dx sin^5x = sec^2 xdx sin^5x-5 dx sin^5x By IBP: = tanx sin^5x- (- 5 sin^6x) cosx tanx dx-5 dx sin^5x = tanx sin^5x+C So f (x )= tanx sin^5x f ( 6 )-f ( 4 )= 2^5 3- ( 2 )^5= 32 3-4 2= 4 3 (8- 6 )

$If \sum (\frac{1 - 5 cos ^{2} x}{sin ^{5} x cos ^{2} x}) dx = f (x) + C, where C is the constant of integration, then f (\frac{π}{6}) - f (\frac{π}{4}) is equal to$

Solution

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