What is the integral of sin^7(2x) ?

The integral of sin^7(2x) is 1/2 (-cos(2x)+cos^3(2x)-(3cos^5(2x))/5+(cos^7(2x))/7)+C

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integral of sin^7(2x)

\int sin^{7} (2 x) d x

\frac{1}{2} (- cos (2 x) + cos^{3} (2 x) - \frac{3 cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7}) + C

Solution steps

\int sin^{7} (2 x) d x

Apply u-substitution

: \int sin^{7} (u) \frac{1}{2} d u

= \int sin^{7} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sin^{7} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - cos^{2} (u))^{3} sin (u) d u

Apply u-substitution

: \int - (1 - v^{2})^{3} d v

= \frac{1}{2} \cdot \int - (1 - v^{2})^{3} d v

Expand

- (1 - v^{2})^{3} : - 1 + 3 v^{2} - 3 v^{4} + v^{6}

= \frac{1}{2} \cdot \int - 1 + 3 v^{2} - 3 v^{4} + v^{6} d v

Apply the Sum Rule:

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

= \frac{1}{2} (- \int 1 d v + \int 3 v^{2} d v - \int 3 v^{4} d v + \int v^{6} d v)

\int 1 d v = v

\int 3 v^{2} d v = v^{3}

\int 3 v^{4} d v = \frac{3 v ^{5}}{5}

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

= \frac{1}{2} (- v + v^{3} - \frac{3 v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= \frac{1}{2} (- cos (2 x) + cos^{3} (2 x) - \frac{3 cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7})

Add a constant to the solution

= \frac{1}{2} (- cos (2 x) + cos^{3} (2 x) - \frac{3 cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7}) + C

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What is the integral of sin^7(2x) ?
The integral of sin^7(2x) is 1/2 (-cos(2x)+cos^3(2x)-(3cos^5(2x))/5+(cos^7(2x))/7)+C