What is the integral of sqrt(sin(x))cos^3(x) ?

The integral of sqrt(sin(x))cos^3(x) is 2/3 sin^{3/2}(x)-2/7 sin^{7/2}(x)+C

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integral of sqrt(sin(x))cos^3(x)

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

\frac{2}{3} sin^{\frac{3}{2}} (x) - \frac{2}{7} sin^{\frac{7}{2}} (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u (1 - u^{2}) : u - u^{\frac{5}{2}}

= \int u - u^{\frac{5}{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

= \int u d u - \int u^{\frac{5}{2}} d u

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

= \frac{2}{3} u^{\frac{3}{2}} - \frac{2}{7} u^{\frac{7}{2}}

Substitute back

u = sin (x)

= \frac{2}{3} sin^{\frac{3}{2}} (x) - \frac{2}{7} sin^{\frac{7}{2}} (x)

Add a constant to the solution

= \frac{2}{3} sin^{\frac{3}{2}} (x) - \frac{2}{7} sin^{\frac{7}{2}} (x) + C

y^{^{'}} + 6 (tan (6 x)) y = - 2 cos (6 x)

\frac{d y}{d x} = \frac{x}{sin ( y )} - \frac{9}{sin ( y )}

\int_{1}^{2 y} \frac{y}{x} d x

3 cos (2 x)

x \to 2 lim (6 x - 2)

What is the integral of sqrt(sin(x))cos^3(x) ?
The integral of sqrt(sin(x))cos^3(x) is 2/3 sin^{3/2}(x)-2/7 sin^{7/2}(x)+C