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Question Number 99779 by john santu last updated on 23/Jun/20
Commented by Dwaipayan Shikari last updated on 23/Jun/20
∫tan6xsec2xsec2xdx=∫t6(1+t2)dt=t77+t99+constant=tan7x7+tan9x9+C{supposetanxast}
Commented by bemath last updated on 23/Jun/20
∫tan6x(1+tan2x)d(tanx)=∫tan6x+tan8xd(tanx)=17tan7x+19tan9x+c
Answered by mathmax by abdo last updated on 23/Jun/20
I=∫tan6xcos4xdxwehave1+tan2x=1cos2x⇒1cos4x=(1+tan2x)2⇒I=∫tan6x(1+tan2x)2dx=tanx=t∫t6(1+t2)21+t2dt=∫t6(1+t2)dt=∫t6dt+∫t8dt=t77+t99+C=tan7x7+tan9x9+C
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