Question

$\int \frac{5^{\cot x}}{{\sin }^{2}x}dx=$

• A. $\frac{-5^{\cot x}}{\log 5}+c$
• B. $\frac{5^{\cot x}}{\log 5}+c$
• C. $5^{\cot x}+c$
• D. ${}_{-}{5}^{\cot x}+c$

Answer 1

The correct option is A $\frac{-5^{\cot x}}{\log 5}+c$

$\int \frac{5^{cotx}}{si{n}^{2}x}dx=\int 5^{cotx}\cdot cose{c}^{2}xdx$

$d\left(cotx\right)=-coses{c}^{2}x\cdot dx$

$cose{c}^{2}x\cdot dx=-dcotx$

$-\int 5^{cotx}d\left(cotx\right)$

Let $cotx=p$

$=-\int 5^{p}dp=-\frac{5^p}{log5}+c$

$\int \frac{5^{cotx}}{sin2x}dx=\frac{-5^{cotx}}{log5}+c$