Question

5⁴ˣ⁺³-3sin(2x+3),Integrate the given function defined on proper domain w.r.t. x.

Answer 1

Dear Student,

Answer:∫ 5⁴ˣ⁺³-3sin(2x+3) dx =
$\frac{1}{4 \: log5} ( {5}^{4x + 3}) + \frac{3}{2} cos(2x + 3) + C$

Solution:

∫ 5⁴ˣ⁺³-3sin(2x+3) dx =
∫${5}^{4x} \times {5}^{3} dx - 3sin(2x + 3)dx$
formula used

∫${a}^{x} dx = \frac{ {a}^{x} }{log \: a} + C$

∫$sin \: x \: dx = - cos \: x \: + C$

$= {5}^{3} \frac{ {5}^{4x} }{4 \: log(5) } - 3( - \frac{1}{2} \cos(2x + 3)) + C \\ \\ = \frac{ {5}^{4x + 3} }{4 \: log(5) } + \frac{3}{2} \cos(2x + 3) + C$

Hope it helps you.

Answer 2

we have to find $\int{5^{4x+3}-3sin(2x+3)}\,dx$

we know, $\int{A^{(ax+b)}}\,dx=\frac{A^{(ax+b)}}{alogA}+C$

and $\int{sin(ax+b)}\,dx=-\frac{1}{a}cos(ax+b)+C$

now, $\int{5^{4x+3}-3sin(2x+3)}\,dx$

= $\int{5^{4x+3}}\,dx-3\int{sin(2x+3)}\,dx$

= $\frac{1}{4log5}5^{4x+3}-3\frac{\{-cos(2x+3)\}}{2}+C$

= $\frac{1}{4log5}e^{4x+3}+\frac{3}{2}cos(2x+3)+C$