# Evaluate the following integrals. intfrac{4^{cot x}}{sin^2x}dx

Question
Integrals
Evaluate the following integrals.
$$\int\frac{4^{\cot x}}{\sin^2x}dx$$

2020-11-21
$$\text{Given integral is }\int\frac{4^{\cot x}}{\sin^2x}$$
$$\text{Rewrite the above integrtal as }\int\frac{4^{\cot x}}{\sin^2x}=\int4^{\cot x}\csc^2(x)dx$$
$$\text{Consider }u=\cot x.$$
$$\text{Then,}$$
$$\frac{du}{dx}=-\csc^2x$$
$$dx=-\frac{du}{\csc^2x}$$
$$\text{Substitute }u=\cot x\ and\ dx=-\frac{du}{\csc^2x}\ in\ \int4^{\cot x}\csc^2(x)dx.$$
$$\int4^{\cot x}\csc^2(x)dx=\int4^u\csc^2(x)(-\frac{du}{\csc^2x})$$
$$=-\int4^udu$$
$$=\frac{-4^u}{\ln(4)}+C$$
$$=-\frac{4^{\cot x}}{\ln(4)}+C$$
$$=-\frac{2^{2\cot x}}{\ln(2)^2}+C$$
$$=-\frac{2^{2\cot x}}{2\ln(2)}+C$$
$$=-\frac{2^{2\cot x-1}}{\ln(2)}+C$$

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