1.

Find the integral pf sin 3x​

Answer»

\implies \displaystyle \int \sf <klux>SIN</klux> \: 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


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