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

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

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integral of 7cos^7(x)

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

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

Solution steps

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

Rewrite using trig identities

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

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

Simplify

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

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

Add a constant to the solution

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

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

\int 7 + cos (x) d x

\int 7 sin^{4} (x) cos (x) d x

\int \frac{3 x - 5}{x ^{2} + 4 x} d x

\int sinh (x) + sinh^{3} (x) d x

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