Question

evaluate integral of cos(3z+4).dz​

Answer 1

$\implies \displaystyle \int \bf \cos(3z + 4 )\: dz$

We will solve the above integration problem by using substitution method :-

$\bf let \: (3z + 4 )\: = t$

$\bf \:dz = \dfrac{dt}{3}$

$\implies \displaystyle \int \bf \cos(t)\: \dfrac{dt}{3}$

$\implies \dfrac{1}{3} \displaystyle \int \bf \cos(t)\: {dt}$

We know that :-

$\underline {\boxed {\displaystyle \int \bf \cos(x)\: {dx= sin \: x+ c}}}$

$\implies \dfrac{1}{3} \displaystyle \int \bf \cos(t) {dt} = \frac{1}{3} sin t+ c$

Now put t = 3z+4

$\implies \displaystyle \bf \frac{1}{3} sin (3z + 4)+ c$

Answer :

$\displaystyle \int \bf \cos(3z + 4 )\: dz = \frac{1}{3} sin (3z + 4)+ c$

Where c is constant