Question

integration of sin x upon sin 3x​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

sin 3x = 3 sin x - 4 sin^3 x

sin x/sin 3x = 1/(3–4sin^2x)

so for integration of 1/(3–4sin^2x ), multiply & divide numerator & denominator by sec^2 x , we get

sec^2 x dx / (3 sec^2 x-4 tan^2 x) = sec^2 x dx / (3(1+tan^2 x)

= sec^2 x dx / (3–tan^2x)

now put tan^2 x = t

which gives sec^2x dx = dt , so

we get integration of dt / (3-t^2) = 1/ (2*square root of (3) * log | (square root of 3 +t )/ square root of 3 - t |+c

= square root of 3 / 6 log | ((square root of 3 )+ tan x )/ ( square root of 3 |+c