# a. A particle moves on a straight line with velocity function v\left ( t \right ) \ = \ sin \...

## Question:

a. A particle moves on a straight line with velocity function {eq}v\left ( t \right ) \ = \ sin \ \omega t \ cos^{2}\omega t {/eq}. Find its position function s = f(t) if f(0) = 0.

s(t) = _____

b. Evaluate the integral using the indicated trigonometric substitute.

{eq}\int \frac{x^{3}}{\sqrt{x^{2} \ + \ 36}} \ dx, \ \ x \ = \ 6 \ tan \ \theta {/eq}

## Trigonometric Substitution in Integration

{eq}{/eq}

Many times while solving an integral, we come across situations where a trigonometric substitution would make the integrand a lot easier to integrate. Therefore, if we know the interval in which our integration variable (say, x) lies, we can apply suitable substitution to proceed with simpler steps.

{eq}1. \ -1 \leq x \leq 1 \\ Put \ x = \sin{\theta } \quad or \quad x = \cos{\theta} \\ {/eq}

{eq}2. \ x \ \in \ R - ( -1 , 1 ) \\ Put \ x = \sec{\theta } \quad or \quad x = \csc{\theta} \\ {/eq}

{eq}3. \ x \ \in \ R \\ Put \ x = \tan{\theta } \quad or \quad x = \cot{\theta} \\ {/eq}

{eq}4. \ a \leq x \leq b \\ Put \ x = a\cos^2{\theta} + b \sin^2{\theta} \\ {/eq}

The above substitutions prove to be very useful while simplifying the integrals involving nested/rational functions.

{eq}{/eq}

{eq}{/eq}

{eq}\displaystyle{ (A) \ v(t) = \sin{\omega t} \cos^2{\omega t} \\ \Rightarrow \frac{ds}{dt} = \sin{\omega t} \cos^2{\omega t} \\ \Rightarrow ds = \sin{\omega t} \cos^2{\omega t} \ dt \\ \Rightarrow \int ds = \int \sin{\omega t} \cos^2{\omega t} \ dt \\ } {/eq}

{eq}\displaystyle{ Let, \ \cos{\omega t} = x \\ \Rightarrow dx = - \omega \sin{\omega t} \ dt \\ \Rightarrow \sin{\omega t} \ dt = -\frac{dx}{\omega} \\ } {/eq}

Substituting these values in the integral, we get :

{eq}\displaystyle{ \int ds = - \int x^2 \ \frac{dx}{\omega} \\ \Rightarrow s = -\frac{1}{\omega} \int x^2 \ dx \\ \Rightarrow s = -\frac{x^3}{3\omega} + c \\ \Rightarrow s = -\frac{\cos^3{\omega t}}{3\omega} + c \\ } {/eq}

c is a constant of integration which can be evaluated by the given initial condition ( s = 0 for t = 0 )

{eq}\displaystyle{ \therefore -\frac{\cos^3{\omega (0)}}{3\omega} + c = 0 \\ \Rightarrow c - \frac{1}{3\omega} = 0 \\ \Rightarrow c = \frac{1}{3\omega} \\ } {/eq}

$$\displaystyle s = \frac{1- \cos^3{\omega t}}{3\omega} \\$$

{eq}\displaystyle{ (B) \ I = \int \frac{x^{3}}{\sqrt{x^{2} + 36}} \ dx \quad x = 6 \tan{\theta} \\ \Rightarrow dx = 6 \sec^2{\theta} d\theta \\ } {/eq}

{eq}\displaystyle{ \therefore I = \int \frac{(6\tan{\theta})^3}{\sqrt{36\tan^2{\theta} + 36}} \ 6 \sec^2{\theta} d\theta \\ \Rightarrow I = \int \frac{1296\tan^3{\theta} \sec^2{\theta} }{6\sec{\theta}} \\ \Rightarrow I =216 \int \tan^3{\theta} \sec{\theta} d\theta \\ \Rightarrow I = 216 \int \tan^2{\theta} \sec{\theta} \tan{\theta} d\theta \\ \Rightarrow I = 216 \int (\sec^2{\theta} -1 ) \sec{\theta} \tan{\theta} d\theta \\ } {/eq}

{eq}Let, \ \sec{\theta} = t \\ \Rightarrow dt = \sec{\theta} \tan{\theta} d\theta \\ {/eq}

Putting these values in the integral, we get :

{eq}\displaystyle{ I =216 \int (t^2 - 1)dt \\ \Rightarrow I = 216(\frac{t^3}{3} - t) + c \\ \Rightarrow I = 216(\frac{\sec^3{\theta}}{3} - \sec{\theta}) + c \\ \Rightarrow I = 216 (\frac{(1+\tan^2{\theta})^{\frac{3}{2}}}{3} - \sqrt{1+\tan^2{\theta}} ) + c \\ \Rightarrow I = 216 (\frac{(1+\frac{x^2}{36})^{\frac{3}{2}}}{3} - \sqrt{1+\frac{x^2}{36}}) + c \\ \Rightarrow I = 216 (\frac{(x^2 + 36)^{\frac{3}{2}}}{648} - \frac{\sqrt{x^2 + 36}}{6}) + c \\ \Rightarrow I = \frac{(x^2 + 36)^{\frac{3}{2}}}{3} - 36\sqrt{x^2 + 36} + c} {/eq}

c is the constant of integration.