Question

Solve this :
$\displaystyle \rm \int \sin(3x + 2)dx$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle \rm \int \sin(3x + 2)dx$

To evaluate this integral, we use method of Substitution

So, Substitute

$\rm \: 3x + 2 = y$

$\rm \: \dfrac{d}{dx}( 3x + 2) =\dfrac{d}{dx} y$

$\rm \: 3dx = dy$

$\rm \: dx \: = \: \dfrac{dy}{3}$

So, on substituting these values in above integral, we get

$\rm \: = \: \displaystyle \rm \int siny \: \frac{dy}{3}$

$\rm \: = \: \dfrac{1}{3}\displaystyle \rm \int siny \: dy$

$\rm \: = \: - \: \dfrac{1}{3}cosy \: + \: c$

$\rm \: = \: - \: \dfrac{1}{3}cos(3x + 2) \: + \: c$

Hence,

$\rm\implies\boxed{{\rm \displaystyle \rm \int \sin(3x + 2)dx = - \dfrac{1}{3}cos(3x + 2) + c }}$

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Short Cut Trick :-

$\boxed{\sf{ \:\displaystyle \rm \int sin(ax + b)dx \: = \: - \: \frac{cos(ax + b)}{a} + c \: }}$

So, using this trick,

$\rm\implies\boxed{{\rm \displaystyle \rm \int \sin(3x + 2)dx = - \dfrac{cos(3x + 2)}{3} + c }}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

$\rm - \frac{1}{3} \cos(3x + 2) + c$

Step-by-step explanation:

To Solve :

$\rm⇢ \: \: \rm \displaystyle \int \sin(3x + 2)dx$

Solution :

Let's substitute t = 3x + 2 ,

⇒ 3dx = dt

then,

$\rm⇢ \: \: \int \sin(t) \frac{dt}{3}$

$\rm⇢ \: \: \frac{1}{3} \int \sin(t)dt$

$\rm⇢ \: \: \frac{1}{3} [ - \cos(t)] + c$

$\bf⇢ \: \: - \frac{ 1}{3} \cos(3x + 2) + c$

FINAL ANSWER :

$\rm \frac{ 1}{3} \cos(3x + 2) + c$