# Integrate \int \cos^2 x \ sm^3x \ dx Solve y=\frac {\cos 2x}{\sin 3y}

## Question:

1. Integrate {eq}\displaystyle \int \cos^2 x \sin^3x \ dx {/eq}

2. Solve {eq}\displaystyle y'=\frac {\cos 2x}{\sin 3y} {/eq}

## U-Substitution

The method of substitution or u-substitution is a technique applicable in integration. In this method, a new variable will be introduced to the integrand so that we can rewrite the integral into a very simple integral. It is the reverse of chain rule in differentiation.

## Answer and Explanation:

**1.**

We can rewrite the given interval as

{eq}\displaystyle \int \cos^2 x \sin^3x \ dx = \int \cos^2 x (\sin^2x )(\sin x) \ dx {/eq}

Apply the trigonometric identity: {eq}\sin^2 \theta + \cos^2 \theta = 1 {/eq}.

{eq}\displaystyle = \int \cos^2 x (1- \cos^2 x )(\sin x) \ dx \\ \displaystyle = \int (\cos x)^2 (1- (\cos x)^2 )(\sin x) \ dx {/eq}

We can now apply the method of substitution. Let

{eq}u = \cos x \\ du = -\sin x \,dx \to dx = -\dfrac{du}{\sin x} {/eq}

The integral becomes

{eq}\displaystyle = \int (u)^2 (1- (u)^2 )(\sin x) \left ( -\dfrac{du}{\sin x} \right ) \\ \displaystyle = \int -u^2 (1- u^2 )\,du \\ \displaystyle = \int (-u^2 + u^4 )\,du {/eq}

Apply sum rule: {eq}\int (f(x)\pm g(x))\,dx = \int f(x)\,dx \pm \int g(x)\,dx {/eq}.

{eq}\displaystyle = \int -u^2 \,du + \int u^4 \,du \\ \displaystyle = -\int u^2 \,du + \int u^4 \,du {/eq}

Apply power rule: {eq}\int x^n\,dx = \frac{x^{n+1}}{n+1} {/eq}.

{eq}\displaystyle = -\left ( \frac{u^{2+1}}{2+1} \right ) + \left ( \frac{u^{4+1}}{4+1} \right ) \\ \displaystyle = -\left ( \frac{u^{3}}{3} \right ) + \left ( \frac{u^{5}}{5} \right ) \\ \displaystyle = - \frac{u^{3}}{3} + \frac{u^{5}}{5} {/eq}

Substitute back {eq}u = \cos x {/eq}.

{eq}\displaystyle = - \frac{(\cos x)^{3}}{3} + \frac{(\cos x)^{5}}{5} \\ \displaystyle = - \frac{\cos^{3} x}{3} + \frac{\cos^{5} x}{5} {/eq}

Finally, add a constant of integration.

{eq}\displaystyle \boldsymbol{= - \frac{\cos^{3} x}{3} + \frac{\cos^{5} x}{5} + C} {/eq}

**2.**

{eq}\displaystyle y'=\frac {\cos (2x)}{\sin (3y)} \\ \displaystyle \frac{dy}{dx} = \frac {\cos (2x)}{\sin (3y)} {/eq}

We can separate the variables.

{eq}\displaystyle dy = \frac {\cos (2x)}{\sin (3y)}\,dx \\ \displaystyle \sin (3y)\, dy =\cos (2x) \,dx {/eq}

Integrate both sides.

{eq}\displaystyle \int \sin (3y)\, dy = \int \cos (2x) \,dx {/eq}

To solve this integrals, we can use u-substitution to both sides. Let

{eq}u = 3y \to du = 3\,dy \to dy = \dfrac{du}{3} \\ v = 2x \to dv = 2\,dx \to dx = \dfrac{dv}{2} {/eq}

The integrals become

{eq}\displaystyle \int \sin (u)\, \left ( \frac{du}{3} \right ) = \int \cos (v) \,\left ( \frac{dv}{2} \right ) \\ \displaystyle \int \frac{\sin (u)}{3}\, du = \int \frac{\cos (v)}{2} \,dv {/eq}

Take out the constants.

{eq}\displaystyle \frac{1}{3} \int \sin (u)\, du = \frac{1}{2} \int \cos (v) \,dv \\ \displaystyle \frac{1}{3} (-\cos(u)) = \frac{1}{2} (\sin (v)) \\ \displaystyle -\frac{\cos(u)}{3} = \frac{\sin(v)}{2} {/eq}

Substitute back the values of {eq}u {/eq} and {eq}v {/eq}.

{eq}\displaystyle -\frac{\cos(3y)}{3} = \frac{\sin(2x)}{2} \\ \displaystyle 6\left [-\frac{\cos(3y)}{3} \right ] = \left [\frac{\sin(2x)}{2} \right ]6 \\ \displaystyle -2\cos(3y) = 3\sin(2x) {/eq}

Add a constant of integration on the side of the {eq}x {/eq}-variable.

{eq}\displaystyle -2\cos(3y) = 3\sin(2x) + C {/eq}

Write the solution as {eq}y {/eq} in terms of {eq}x {/eq}.

{eq}\displaystyle \cos(3y) = \frac{3\sin(2x) + C}{-2} \\ \displaystyle \cos(3y) = -\frac{3}{2}\sin(2x) + C \\ \displaystyle 3y = \arccos \left (-\frac{3}{2}\sin(2x) + C \right ) {/eq}

The solution is

{eq}\displaystyle \boldsymbol{y = \frac{\arccos \left (-\frac{3}{2}\sin(2x) + C \right )}{3}} {/eq}