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

The integral of 7sin^7(x) is -7cos(x)+7cos^3(x)-21/5 cos^5(x)+cos^7(x)+C

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integral of 7sin^7(x)

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

- 7 cos (x) + 7 cos^{3} (x) - \frac{21}{5} cos^{5} (x) + cos^{7} (x) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= 7 \cdot \int (1 - cos^{2} (x))^{3} sin (x) d x

Apply u-substitution

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

= 7 \cdot \int - (1 - u^{2})^{3} d u

Expand

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

= 7 \cdot \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

= 7 (- \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}

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

Substitute back

u = cos (x)

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

Simplify

7 (- cos (x) + cos^{3} (x) - \frac{3 cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7}) : - 7 cos (x) + 7 cos^{3} (x) - \frac{21}{5} cos^{5} (x) + cos^{7} (x)

= - 7 cos (x) + 7 cos^{3} (x) - \frac{21}{5} cos^{5} (x) + cos^{7} (x)

Add a constant to the solution

= - 7 cos (x) + 7 cos^{3} (x) - \frac{21}{5} cos^{5} (x) + cos^{7} (x) + C

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What is the integral of 7sin^7(x) ?
The integral of 7sin^7(x) is -7cos(x)+7cos^3(x)-21/5 cos^5(x)+cos^7(x)+C