Calculate the integral of y=sin6x/(5+cos12x).

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neela | High School Teacher | (Level 3) Valedictorian

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To find the integral of y=sin6x/(5+cos12x).

We know that cos2A = 2cos^2A =1.

Therefore cos12A = 2cos^2(6x) - 1.

Therefore y =  sin6x/{5+2cos^2(6x) -1)} = sin6x/{2sin^2(6x)+4}  = sin6x/2{cos^2(6x)+2}

Put cos6x = t, Then  by differentiation we get: -sinx dx = dt. Or sinx dx = Int {1/(t^2+2)-}dt/2.

Int y dx = (-1/2) Int dt/(t^2+2) =  (-1/2) * (1/sqrt2) arc tan (t/sqrt2).

Therefore Int{ sin6x/(5+sin12x) } dx = (-1/2sqrt2) arc tan {cos 6x)/sqrt2} + C.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll write the term from denominator cos 12x = cos 2*(6x)

We'll apply the formula:

cos 2a = 2(cos a)^2 - 1

cos 2*(6x) = 2(cos 6x)^2 - 1

The denominator will become:

5+cos12x = 5 + 2(cos 6x)^2 - 1

5+cos12x = 2(cos 6x)^2 + 4

2(cos 6x)^2 = 2[(cos 6x)^2 + 2]

Int f(x)dx = (1/2)Int sin 6xdx/[(cos 6x)^2 + 2]

We'll substitute cos 6x  = t.

We'll differentiate both sides:

- 6 sin 6x dx = dt

sin 6x dx = -dt/6

We'll write the integral in t:

Int f(x)dx = (-1/12)Int dt/(t^2 + 2) 

 (-1/12)Int dt/(t^2 + 2)  = -(1/12*sqrt2)*arctan t/sqrt2 + C

Int f(x)dx = -(1/12*sqrt2)*arctan (cos 6x)/sqrt2 + C

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