What is the integral of 1+sin^2(x)csc(x) ?

The integral of 1+sin^2(x)csc(x) is x-cos(x)+C

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integral of 1+sin^2(x)csc(x)

\int 1 + sin^{2} (x) csc (x) d x

x - cos (x) + C

Solution steps

\int 1 + sin^{2} (x) csc (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x + \int sin^{2} (x) csc (x) d x

\int 1 d x = x

\int sin^{2} (x) csc (x) d x = - cos (x)

= x - cos (x)

Add a constant to the solution

= x - cos (x) + C

\int \frac{1}{x x + 1} d x

\int (3 cos (x) - 4 sin (x) + 5 sec^{2} (x)) d x

\int sec^{4} (y) tan^{3} (y) d y

\int \frac{2 x ^{2} + x ^{4}}{- 4 x ^{2}} d x

\int \frac{x}{( x ^{2} x ^{2} + 1 )} d x

What is the integral of 1+sin^2(x)csc(x) ?
The integral of 1+sin^2(x)csc(x) is x-cos(x)+C