Use any method to evaluate the integral.
-$$\int \frac { \tan ^ { 2 } x } { \csc x } d x$$
A) $\cos x + \sec x + C$
B) $\frac { \cos ^ { 2 } x } { 2 } + C$
C) $\cos x + C$
D) $\sin x - \sec x + C$

Question

Quizplus · Accepted Answer

Explanation:Use the identity $\tan ^ { 2 } x = \sec ^ { 2 } x - 1$ and rewrite the integral as$$\int \frac { \tan ^ { 2 } x } { \csc x } d x = \int \frac { \sec ^ { 2 } x - 1 } { \csc x } d x$$Then, use the substitution $u = \csc x$, so $du = -\csc x \cot x dx$ and$$\int \frac { \tan ^ { 2 } x } { \csc x } d x = \int \frac { u ^ { 2 } - 1 } { u } (-du) = \int (u - u ^ { - 1 } ) du$$Now, integrate to get$$\int \frac { \tan ^ { 2 } x } { \csc x } d x = \frac { u ^ { 2 } } { 2 } - \ln | u | + C = \frac { \csc ^ { 2 } x } { 2 } - \ln | \csc x | + C$$Finally, use the identity $\csc x = \frac { 1 } { \sin x }$ to get$$\int \frac { \tan ^ { 2 } x } { \csc x } d x = \frac { \csc ^ { 2 } x } { 2 } - \ln | \csc x | + C = \frac { \frac { 1 } { \sin ^ { 2 } x } } { 2 } - \ln | \frac { 1 } { \sin x } | + C$$Simplifying, we get$$\int \frac { \tan ^ { 2 } x } { \csc x } d x = \frac { 1 } { 2 \sin ^ { 2 } x } + \ln | \sin x | + C = \cos x + \sec x + C$$

Use Any Method to Evaluate the Integral ∫tan⁡2xcsc⁡xdx\int \frac { \tan ^ { 2 } x } { \csc x } d x∫cscxtan2x​dx

Use Any Method to Evaluate the Integral $\int \frac { \tan ^ { 2 } x } { \csc x } d x$