What is the integral of (sin^2(2x))/(csc^3(x)) ?

The integral of (sin^2(2x))/(csc^3(x)) is 4(-(cos^3(x))/3+(2cos^5(x))/5-(cos^7(x))/7)+C

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integral of (sin^2(2x))/(csc^3(x))

\int \frac{sin ^{2} ( 2 x )}{csc ^{3} ( x )} d x

4 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7}) + C

Solution steps

\int \frac{sin ^{2} ( 2 x )}{csc ^{3} ( x )} d x

Use the basic trigonometric identity:

\frac{1}{csc ( x )} = sin (x)

= \int sin^{3} (x) sin^{2} (2 x) d x

Rewrite using trig identities

= \int 4 sin^{5} (x) cos^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int sin^{5} (x) cos^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int (1 - cos^{2} (x))^{2} sin (x) cos^{2} (x) d x

Apply u-substitution

: \int - u^{2} (1 - u^{2})^{2} d u

= 4 \cdot \int - u^{2} (1 - u^{2})^{2} d u

Expand

- u^{2} (1 - u^{2})^{2} : - u^{2} + 2 u^{4} - u^{6}

= 4 \cdot \int - u^{2} + 2 u^{4} - u^{6} d u

Apply the Sum Rule:

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

= 4 (- \int u^{2} d u + \int 2 u^{4} d u - \int u^{6} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

\int u^{6} d u = \frac{u ^{7}}{7}

= 4 (- \frac{u ^{3}}{3} + \frac{2 u ^{5}}{5} - \frac{u ^{7}}{7})

Substitute back

u = cos (x)

= 4 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7})

Add a constant to the solution

= 4 (- \frac{cos ^{3} ( x )}{3} + \frac{2 cos ^{5} ( x )}{5} - \frac{cos ^{7} ( x )}{7}) + C

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What is the integral of (sin^2(2x))/(csc^3(x)) ?
The integral of (sin^2(2x))/(csc^3(x)) is 4(-(cos^3(x))/3+(2cos^5(x))/5-(cos^7(x))/7)+C