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Mathematics 7 Online
OpenStudy (anonymous):

solve the integral. (or find y) y'=sin2x/(1+cos^2x)

OpenStudy (abb0t):

Is this algebra 2A?

OpenStudy (anonymous):

no its lvl 2 Calculus.

OpenStudy (anonymous):

\[\int\frac{\sin2x}{1+\cos^2x}~dx~~?\]

OpenStudy (anonymous):

\(\sin2x=2\sin x\cos x\), so you have \[2\int\frac{\sin x\cos x}{1+\cos^2x}~dx\] Substitute \(u=\cos x\), so that \(-du=\sin x~dx\): \[-2\int\frac{u}{1+u^2}~du\] Make another substitution: \(t=1+u^2\), so \(\dfrac{1}{2}dt=u~du\): \[-\int\frac{dt}{t}\]

OpenStudy (anonymous):

ty

OpenStudy (anonymous):

yw

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