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

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

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integral of cos^7(x)

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

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

Solution steps

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

Rewrite using trig identities

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (ln (4 + x^{2} y^{2}))

(x^{2} + y^{2})^{2} = 4 x^{2} y, (1, 1)

\int \frac{- 12}{x ^{2} 4 - x ^{2}} d x

\frac{\partial}{\partial x} (4 x^{2} + y^{2} - 7 y)

\frac{d}{d x} (\frac{6 x}{x ^{2} + 3})

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