What is the integral of sin^7(y) ?

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

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integral of sin^7(y)

\int sin^{7} (y) d y

- cos (y) + cos^{3} (y) - \frac{3 cos ^{5} ( y )}{5} + \frac{cos ^{7} ( y )}{7} + C

Solution steps

\int sin^{7} (y) d y

Rewrite using trig identities

= \int (1 - cos^{2} (y))^{3} sin (y) d y

Apply u-substitution

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

= \int - (1 - u^{2})^{3} d u

Expand

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

= \int - 1 + 3 u^{2} - 3 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

= - \int 1 d u + \int 3 u^{2} d u - \int 3 u^{4} d u + \int u^{6} d u

\int 1 d u = u

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

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

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

= - u + u^{3} - \frac{3 u ^{5}}{5} + \frac{u ^{7}}{7}

Substitute back

u = cos (y)

= - cos (y) + cos^{3} (y) - \frac{3 cos ^{5} ( y )}{5} + \frac{cos ^{7} ( y )}{7}

Add a constant to the solution

= - cos (y) + cos^{3} (y) - \frac{3 cos ^{5} ( y )}{5} + \frac{cos ^{7} ( y )}{7} + C

\int \frac{3}{35} x^{2} d x

y^{^{'''}} + 2 y^{^{''}} - 13 y^{^{'}} + 10 y = 0

y^{^{''}} + 2 y^{^{'}} + 4 y = 0

\int \frac{7}{1 + x ^{2}} d x

x^{2} ln (7 x)

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