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

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

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integral of sin^7(x)

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

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

Solution steps

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

Rewrite using trig identities

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

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 (x)

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

Add a constant to the solution

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

\frac{d}{d x} (ln (x + 4))

\int \frac{x ^{2}}{( 2 x + 1 ) ^{5}} d x

\frac{d}{d x} (\frac{x}{1 + e ^{- x}})

\int sin^{3} (8 x) d x

[x + (x + sin^{2} (x))^{5}]^{3}

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