Given integral can be written as
I=∫(tanx)34sec2xdx
Let tanx=t
⇒sec2xdx=dt
⇒I=∫t−34dt
Using ∫xndx=n+1xn+1+C, we get
I=(−31)t−31+C
⇒I=−t313+C
⇒I=−3tan−31x+C.
∫sec2x⋅cot34xdx is equal to
Held on 9 Apr 2019 · Verified 6 Jul 2026.
3tan−31x+C
−43tan−34x+C
−3tan−31x+C
−3cot−31x+C
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