Question

$\int \frac{7^{2x+3}{\sin }^{2}2x+{\cos }^{2}2x}{{\sin }^{2}2x}=\frac{7^{2x+3}}{2\log 7}-\frac{(\cot x+x)}{b}$.Find $b$

Answer 1

$\int [7^{2x+3}+co{t}^2(2x\left)\right]dx=7^3\int 7^{2x}dx+\int \left[cose{c}^2\right(2x)-1]dx=7^3\int {49}^{x}dx-\left(\frac{cot\left(2x\right)}{2}\right)-x+C=7^3\left(\frac{{49}^x}{log49}\right)-\left(\frac{cot\left(2x\right)}{2}\right)-x+C=\left(\frac{7^{2x+3}}{2log7}\right)-\left(\frac{\left[cot\right(2x)+2x]}{2}\right)+C\therefore b=2$