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Question Number 183976 by cortano1 last updated on 01/Jan/23
∫sinx−1+sinxcosx−1+cosxdx=?
Answered by MJS_new last updated on 01/Jan/23
t=1+sinx2cosx2⇔x=2arctant2−12t→dx=4dtt2−1⇒∫sinx−1+sinxcosx−1+cosxdx==4∫t4−2t3+6t−1(t2+1)(t2+2(1+5)t+1)(t2+2(1−5)t+1)dtnowdothedecompositionetc.foryourselfbecauseI′mtoolazyagain.
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