Question

Integration questions .

Answer 1

1)

`\\\\\ \textbf{a)}\\\\~~~\displaystyle \int (6x- \sin 3x) ~ dx\\\\=6\displaystyle \int x ~ dx - \displaystyle \int \sin 3x ~ dx\\\\=6 \cdot \frac{x^2}2 - \frac 13 (- \cos 3x) +C~~~~~~~~~~~;\left[\displaystyle \int x^n~ dx = (x^(n+1))/(n+1)+C,~~~n \\eq -1\right]\\\\ =3x^2 +\frac{\cos 3x}3 +C~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \sin (mx) ~dx = -\frac 1m ~ (\cos mx)+C \right]\\`

`\textbf{b)}\\\\~~~~\displaystyle \int(3e^(-2x) +\cos (0.5 x)) dx\\\\=3\displaystyle \int e^(-2x) ~dx+ \displaystyle \int \cos(0.5 x) ~dx\\\\\\=-\frac 32 e^(-2x) + \frac 1{0.5} \sin (0.5 x) +C~~~~~~~~~~~~~~;\left[\displaystyle \int e^(mx)~dx = \frac 1m e^(mx) +C \right]\\\\\\=-\frac 32 e^(-2x) + 2 \sin(0.5 x) +C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \cos(mx)~ dx = \frac 1m \sin(mx) +C\right]\\\\\\=-1.5e^(-2x) +2\sin(0.5x) +C`

2)

`\textbf{a)}\\\\y = \displaystyle \int \cos(x+5) ~ dx\\\\\text{Let,}\\\\~~~~~~~u = x+5\\\\\implies (du)/(dx) = 1+0~~~~~~;[\text{Differentiate both sides.}]\\\\\implies (du)/(dx) = 1\\\\\implies du = dx\\\\\text{Now,}\\\\y= \displaystyle \int \cos u ~ du\\\\~~~= \sin u +C\\\\~~~=\sin(x+5) + C`

`\textbf{b)}\\\\y = \displaystyle \int 2(5x-3)^4 dx\\\\\text{Let,}\\~~~~~~~~u = 5x-3\\\\\implies (du)/(dx) = 5~~~~~~~~~~;[\text{Differentiate both sides}]\\\\\implies dx = \frac{du}5\\\\\text{Now,}\\\\y = 2\cdot \frac 1 5 \displaystyle \int u^4 ~ du\\\\\\~~=\frac 25 \cdot (u^(4+1))/(4+1) +C\\\\\\~~=\frac 25 \cdot \frac{u^5}5+C\\\\\\~~=(2u^5)/(25)+C\\\\\\~~=(2(5x-3)^5)/(25)+C`

3)

`\textbf{a)}\\\\y = \displaystyle \int xe^(3x) dx\\\\\text{We know that,}\\\\ \displaystyle \int (uv) ~dx = u \displaystyle \int v ~ dx - \displaystyle \int \left[ (du)/(dx) \displaystyle \int ~ v ~ dx \right]~ dx\\\\\text{Let}, u =x~ \text{and}~ v=e^(3x) .\\\\y= \displaystyle \int xe^(3x) ~dx\\\\\\~~= x\displaystyle \int e^(3x) ~ dx - \displaystyle \int \left[(d)/(dx)(x) \displaystyle \int e^(3x)~ dx \right]~ dx\\\\\\`

`=x\displaystyle \int e^(3x)~ dx - \displaystyle \frac 13 \int \left(e^(3x) \right)~ dx\\\\\\=\frac{xe^(3x)}3 - \frac 13 \cdot \frac{ e^(3x)}3+C\\\\\\= (xe^(3x))/(3)- (e^(3x))/(9)+C\\\\\\=(3xe^(3x))/(9)- \frac{e^(3x)}9 + C\\\\\\= \frac 19e^(3x)(3x-1)+C`