Question

find the integral pf sin 3x​

Answer 1

$\implies \displaystyle \int \sf sin \: 3x \: dx$

$\sf \: let \: 3x = t \\ \therefore \sf3dx = dt \\ \sf dx = \dfrac{dt}{3}$

$\implies \displaystyle \int \sf sin \: t \: \frac{dt}{3}$

$\implies \displaystyle \int \sf sin \: t \: \times \frac{1}{3} dt$

$\implies \sf\dfrac{1}{3} \displaystyle \int \sf sin \: t \: dt$

$\implies \sf\dfrac{1}{3} cos \: t \: + c$

where c is constant.

Now put t = 3x

$\implies \sf\dfrac{1}{3} cos \: 3x \: + c$

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Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec²x dx = tan x + C

∫ csc²x dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C