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

The integral of (sin(x)+1/(sin(x)))^2 is 5/2 x-1/4 sin(2x)-cot(x)+C

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

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

\frac{5}{2} x - \frac{1}{4} sin (2 x) - cot (x) + C

Solution steps

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

Rewrite using trig identities

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

Expand

\frac{( sin ^{2} ( x ) + 1 ) ^{2}}{sin ^{2} ( x )} : sin^{2} (x) + 2 + \frac{1}{sin ^{2} ( x )}

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

\int sin^{2} (x) d x = \frac{1}{2} (x - \frac{1}{2} sin (2 x))

\int 2 d x = 2 x

\int \frac{1}{sin ^{2} ( x )} d x = - cot (x)

= \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + 2 x - cot (x)

Simplify

\frac{1}{2} (x - \frac{1}{2} sin (2 x)) + 2 x - cot (x) : \frac{5}{2} x - \frac{1}{4} sin (2 x) - cot (x)

= \frac{5}{2} x - \frac{1}{4} sin (2 x) - cot (x)

Add a constant to the solution

= \frac{5}{2} x - \frac{1}{4} sin (2 x) - cot (x) + C

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What is the integral of (sin(x)+1/(sin(x)))^2 ?
The integral of (sin(x)+1/(sin(x)))^2 is 5/2 x-1/4 sin(2x)-cot(x)+C