Mathematics

# $\displaystyle \int {\dfrac{7^{2x+3}\sin^2 2x+ \cos^22x}{\sin^22x}}=\dfrac{7^{2x+3}}{2\log 7}-\dfrac{(\cot x+x)}{b}$.Find $b$

2

##### SOLUTION

$\\\int \left[7^{2x+3}+cot^2(2x)\right]dx\\=7^3\int 7^{2x}dx+\int [cosec^2(2x)-1]dx\\=7^3\int 49^x dx-(\frac{cot(2x)}{2})-x+C\\=7^3 (\frac{49^x}{log 49})-(\frac{cot(2x)}{2})-x+C\\=(\frac{7^{2x+3}}{2log7})-(\frac{[cot(2x)+2x]}{2})+C\\\therefore\>b=2$

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One Word Medium Published on 17th 09, 2020
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