anonymous
  • anonymous
evaluate the integral: ∫sin⁡2x/√(cos^2 x+16) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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.Sam.
  • .Sam.
\[\int\limits \frac{\sin(2x)}{\cos^2x+16}dx\]
.Sam.
  • .Sam.
Hmm try subbing \[u=\cos^2x+16\] \[du=-2\sin(x)\cos(x)dx\] \[-du=2\sin(x)\cos(x)dx\] From \[\int\limits \frac{\sin(2x)}{\cos^2x+16}dx\] \[=\int\limits \frac{2\sin(x)\cos(x)}{\cos^2x+16}dx\] Substituting gives \[=-\int\limits \frac{1}{u} \, du\]
amriju
  • amriju
there's root in the denom. i think

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.Sam.
  • .Sam.
Like this \[\int\limits \frac{ \sin (2 x)}{\sqrt{ \cos ^2x+16}} \, dx\]
amriju
  • amriju
yep..
amriju
  • amriju
ok...so sin 2x= 2sinx cosx..
amriju
  • amriju
let cos x be t...-sin xdx=dt..
amriju
  • amriju
then u'll find the expression reducing to t/(t^2+16)^1/2..
amriju
  • amriju
now 2tdt=d(t^2)..
amriju
  • amriju
u'll then get d(k)/(k+16)^1/2...k=t^2...
.Sam.
  • .Sam.
You don't have to do that, take the whole thing from the denominator, \[u=\cos^2x+16\] \[-du=2\cos(x)\sin(x)dx\] Which gives \[\int\limits\frac{-du}{\sqrt u}\]
amriju
  • amriju
I didn't mention the constants to be multiplied or divided
amriju
  • amriju
lol yeah
.Sam.
  • .Sam.
Yeah so just integrate and sub back u, done

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