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Evaluate the integral.
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\[\sqrt{1+\sin^{2}x} sinxcosx dx\]
what is the trig id of 1+sin^2x?
sorry \[1+\sin ^{2}x\]
cos^2
is the answer: 1/3(sin^2x+1)^3/2 +c??
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sub in cos^2
s^2 + c^2 = 1 c^2 = 1 - s^2 you sure your post is correct?
yup
was my answer correct?
\[cos^2 \ne \sqrt{1+sin^2}\]
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is the answer: 1/3(sin^2x+1)^3/2 +c??
not if youve posted your question correctly :/
thats what wolframalpha told me :(
and you dont trust the wolf? cause I was assuming you were subbing in cos for sqrt(1+sin^2)
i let u=sin^2x+1 and du=2sinxcosx
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that seems a smart thing to do :)
well thats the answer i got then.
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