Indefinite Integration 3 Question 17

17. f(x)=|secxcosxsec2x+cotxcosecx cos2xcos2xcosec2x 1cos2xcos2x|.

Then, 0π/2f(x)dx=.

(1987,2M)

Analytical & Descriptive Questions

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Answer:

Correct Answer: 17. 15π+3260

Solution:

  1. Given, f(x)=|secxcosxcosecxcotx+sec2x cos2xcos2xcosec2x 1cos2xcos2x| Applying R31cosxR3,

f(x)=cosx|secxcosxcosecxcotx+sec2x cos2xcos2xcosec2x secxcosxcosx|

Applying R1R1R3f(x)

=cosx|00cosecxcotx+sec2xcosxcos2xcos2xcosec2xsecxcosxcosx|

=(cosecxcotx+sec2xcosx)(cos3xcosx)cosx=sin2x+cos3xcos3xsin2xsin2xcos2xcos2xsin2x=sin2xcos3x(1sin2x)=sin2xcos5x0π/2f(x)dx=0π/2(sin2x+cos5x)dx0π/2sinmxcosnxdx=m+12n+222m+n+220π/2f(x)dx=321222+6212272=1/2π2+2π2523212π=π4+815=15π+3260



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