Q.469. Indefinite integration.

Question 469.
Integrate cosec^5 (5x) dx.

Answer 469.
Let 5x = u
=> 5 dx = du
=> Integrand
= (1/5) ∫ cosec^5 u du ... ( 1 )

∫ cosec^3 u du
= ∫ cosecu * cosec^2 u du
Using integration by parts
= cosecu ∫ cosec^2 u du - ∫ [d/du(cosecu) ∫ cosec^2u du] du
= - cosecu cotu - ∫ cosecu cot^2 u du
= - cosecu cotu - ∫ cosecu (cosec^2 u - 1) du
= - cosecu cotu - ∫ cosec^3 u du + ∫ cosecu du
=>
2 ∫ cosec^3 u du
= - cosecu cotu + ln ltan(u/2)l + 2c
=> ∫ cosec^3 u du
= - (1/2) cosecu cotu + (1/2) ln ltan(u/2)l + c ... ( 2 )

Now, ∫ cosec^5 u du
= ∫ cosec^3 u * cosec^2 u du
Integrating by parts,
= cosec^3 u ∫ cosec^2 u du - ∫ [d/du(cosec^3 u) ∫ cosecu du] du
= - cosec^3 u * cotu - ∫ 3 cosec^3 u * cot^2u du
= - cosec^3 u * cotu - 3 ∫ cosec^3u (cosec^2u - 1) du
= - cosec^3 u * cotu - 3 ∫ cosec^5 u du + 3 ∫ cosec^3 u du
=> 4 ∫ cosec^5 u du
= - cosec^3 u * cotu + 3 ∫ cosec^3 u du

[Plugging the value of ∫ cosec^3 u du from ( 2 ) above]
= - cosec^3 u * cotu + 3 [ - (1/2) cosecu cotu + (1/2) ln ltan(u/2)l ] + 4c
=> ∫ cosec^5 u du
= - (1/4) cosec^3 u * cotu - (3/8) cosecu cotu + (3/8) ln ltan(u/2)l ] + c

[Plugging in ( 1 ) above]
=> Integrand
= - (1/20) cosec^3 (5x) * cot(5x) - (3/40) cosec(5x) cot(5x)
   + (3/40) ln ltan(5x/2)l + c'. [c' = c/5]

Confirmation that the above answer is correct as verified by Wolfram Alpha:
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