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Question Number 116501 by bemath last updated on 04/Oct/20
∫sinx−cosxsin2xdx?
Answered by TANMAY PANACEA last updated on 04/Oct/20
ddx(sinx+cosx)=cosx−sinx∫−d(sinx+cosx)1−1+sin2x∫−d(sinx+cosx)(sinx+cosx)2−1∫dtt2−1formula
Commented by bemath last updated on 04/Oct/20
=−ln∣t+t2−1∣+c=−ln∣sinx+cosx+sin2x∣+c?
Commented by MJS_new last updated on 04/Oct/20
yes
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