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Question: Answered & Verified by Expert
If f(x)=cosec5xdx, then fπ4=
MathematicsIndefinite IntegrationAP EAMCETAP EAMCET 2018 (25 Apr Shift 1)
Options:
  • A -14[32-5log(2+1)]+c
  • B -18[52-3log(2+1)]+c
  • C -18[72+3log(2+1)]+c
  • D 18[52+log(2+1)]+c
Solution:
1081 Upvotes Verified Answer
The correct answer is: -18[72+3log(2+1)]+c

Given:

fx=cosec5x dx

        =cosec3xIcosec2xIIdx

=cosec3x-cotx--3cosec3xcotx-cotxdx

=-cosec3xcotx-3cosec3xcot2xdx

=-cosec3xcotx-3cosec3xcosec2x-1dx

=-cosec3xcotx-3cosec5xdx+3cosec3xdx

=-cosec3xcotx-3fx+3cosec3xdx

4fx=-cosec3xcotx+3cosec3xdx

4fx=-cosec3xcotx+3I  ...i

Now,

I=cosec3xdx

  =cosecxcosec2xdx

=cosecxcosec2xdx--cosecx cotx-cotxdx

=-cosecx cotx-cosecx cot2xdx

=-cosecx cotx-cosecxcosec2x-1dx

=-cosecx cotx-cosec3xdx+cosecxdx

I=-cosecx cotx-I+cosecx dx

2I=-cosecx cotx+logcosecx-cotx+C

I=12-cosecx cotx+logcosecx-cotx+C

Putting in i, we get

4fx=-cosec3x cotx+32 -cosecx cotx+logcosecx-cotx+c

Putting x=π4 in above equation, we get

4fπ4=-cosecπ43cotπ4+32-cosecπ4cotπ4+logcosecπ4-cotπ4+c'

 4fπ4=-22+32-2+log2-1+c'

4fπ4=-722+32log2-1+c'

4fπ4=-722+32log2-12+12+1+c'

4fπ4=-722+32log12+1+c'

4fπ4=-722-32log2+1+c'

fπ4=-1872+3log2+1+c.

 

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