Question

7²ˣ⁺³ sin²2x+cos²2x/sin²2x,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

we have to find $\int{\frac{7^{2x+3}sin^22x+cos^22x}{sin^22x}}\,dx$

=$\int{\frac{7^{2x+3}sin^22x}{sin^22x}}\,dx+\int{\frac{cos^22x}{sin^22x}}\,dx$

=$\int{7^{2x+3}}\,dx+\int{cot^22x}\,dx$

= $\int{7^{2x+3}}\,dx+\int{cosec^22x-1}\,dx$

=$\int{7^{2x+3}}\,dx+\int{cosec^22x}\,dx-\int{dx}$

we know,
$\int{A^{(ax+b)}}\,dx=\frac{1}{alogA}a^{(ax+b)}+C$

$\int{cosec^2(ax+b)}\,dx=-\frac{1}{a}cot(ax+b)+C$

=$\frac{1}{2log7}7^{2x+3}+\frac{1}{2}(-cot2x)-x+C$

Answer 2

Hello,

Solution:

$\int{\frac{{7}^{2x + 3} {sin}^{2} 2x + {cos}^{2}2x }{ {sin}^{2}2x }} dx \\ \\ \int{\frac{ {7}^{2x + 3} {sin}^{2}2x }{ {sin}^{2}2x }} dx + \int{\frac{ {cos}^{2}2x }{ {sin}^{2}2x }} dx \\ \\\int{ {7}^{2x + 3}} dx +\int{ {cot}^{2} 2x}dx$
let
$2x + 3 = t \\ 2 \: dx = dt \\ dx = \frac{dt}{2} \\ \frac{1}{2}\int{ {7}^{t}}dt \\ \\ = \frac{1}{2} \frac{ {7}^{t} }{log \: 7} + c \\ \\ undo \: substitution \\ = \frac{{ {7}^{2x + 3} }}{2 \: log7} + c\: \: \: \: \: \: \: ......eq1$
now as we know that
$1 + {cot}^{2} x = {cosec}^{2} x \\ \\{cot}^{2} x = {cosec}^{2} x - 1 \\ \int{cosec}^{2} 2xdx - \int{1}dx \\ = - \frac{1}{2} cot \: 2x - x \: \: \: \: \: ....eq2$

add both equations
$= \frac{{ {7}^{2x + 3} }}{2 \: log7} - \frac{1}{2} cot \: 2x - x + c$

Hope it helps you.