What is the integral of (7cos^5(x))/(sqrt(sin(x))) ?

The integral of (7cos^5(x))/(sqrt(sin(x))) is 7(2sqrt(sin(x))-4/5 sin^{5/2}(x)+2/9 sin^{9/2}(x))+C

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integral of (7cos^5(x))/(sqrt(sin(x)))

\int \frac{7 cos ^{5} ( x )}{sin ( x )} d x

7 (2 sin (x) - \frac{4}{5} sin^{\frac{5}{2}} (x) + \frac{2}{9} sin^{\frac{9}{2}} (x)) + C

Solution steps

\int \frac{7 cos ^{5} ( x )}{sin ( x )} d x

Take the constant out:

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

= 7 \cdot \int \frac{cos ^{5} ( x )}{sin ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{( 1 - u ^{2} ) ^{2}}{u} : \frac{1}{u} - 2 u^{\frac{3}{2}} + u^{\frac{7}{2}}

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

\int \frac{1}{u} d u = 2 u

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

= 7 (2 u - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{9} u^{\frac{9}{2}})

Substitute back

u = sin (x)

= 7 (2 sin (x) - \frac{4}{5} sin^{\frac{5}{2}} (x) + \frac{2}{9} sin^{\frac{9}{2}} (x))

Add a constant to the solution

= 7 (2 sin (x) - \frac{4}{5} sin^{\frac{5}{2}} (x) + \frac{2}{9} sin^{\frac{9}{2}} (x)) + C

\int 5 x ln (6 x) d x

y = x^{3} + 4

\frac{d}{d x} (cos^{2} (sin (5 e^{x} - 1 6^{x})))

\int \frac{1}{t ^{3}} d t

\frac{\partial}{\partial x} ((x) cos (x) cos (y))

What is the integral of (7cos^5(x))/(sqrt(sin(x))) ?
The integral of (7cos^5(x))/(sqrt(sin(x))) is 7(2sqrt(sin(x))-4/5 sin^{5/2}(x)+2/9 sin^{9/2}(x))+C