Evaluate the trig integral?

Question

What is the integral of sin2x/(5sin^2x+2cos^2x) dx

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Answer 1 · 2018-07-19T15:43:52

# - (1/3) ln ( 7 - 3 cos 2x ) + C#

Use #u = cos 2x, gtiving du = - 2 sin 2x dx#. Now,

# int (sin2x)/(5sin^2x+2cos^2x) dx#

#= int ( sin2x )/( 5/2( 1 - cos 2x ) + ( 1 + cos 2x )) dx#

#= int ( 2 sin2x)/( 7 - 3 cos 2x ) dx#

#= int 1/( 7 - 3 u ) du, u in [ -1 1 ]#, and so, #7 - 3 u in [ 4, 10 ]

#= - (1/3) ln ( 7 - 3 u ) + C#

#= - (1/3) ln ( 7 - 3 cos 2x ) + C#

Answer 2 · 2018-07-19T15:44:59

#1/3\ln|3\sin^2x+2|+C#

#\int \frac{\sin 2x}{5\sin^2x+2\cos^2x}\ dx#

#=\int \frac{2\sin x\cosx}{5\sin^2x+2(1-\sin^2x)}\ dx#

#=\int \frac{2\sin x\cosx\ dx}{3\sin^2x+2}\ dx#

Let #\sin x=t\implies \cos x\dx=dt#

#=\int \frac{2t\ dt}{3t^2+2}#

#=2/6\int \frac{6t\ dt}{3t^2+2}#

#=1/3\int \frac{d(3t^2+2)}{3t^2+2}#

#=1/3\ln|3t^2+2|+C#

#=1/3\ln|3\sin^2x+2|+C#