mathematics-class-12-unit-09-chapter-08-differential-equations-8-8-k4kulmitnkm

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Use of differential inequalities 1/4 </primary>
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- Consider the equation
- $\frac{d y}{d x}=y(x+y), \quad y(0)=0$ (4.1)

- We see at once that (4.1) is a Bernoulli equation but we cannot avail of this since solving it as a Bernoulli equation would involve division by $y^2$ and that is illegal in view of the fact that $y(0)-0$.

- We see by inspection that $y(x)=0$ is a solution.

- How do we know that there are no other solutions satisfying $y(0)-0$ ?

- A general theorem on differential equations does guarantee that there are no other solutions but we have not even stated this theorem!

- Recall the idea in the first lecture of knocking off some terms and making the differential equation into an inequality that is easier to handle.
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<footer style = "font-size: 18px"> $\rightarrow$ Differential equations lecture-8 $\rightarrow$ <span style = "color:blue"> Use of differential inequalities 1/4 </span> $\rightarrow$ Use of differential inequalities 2/4 $\rightarrow$ Use of differential inequalities 3/4 </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Use of differential inequalities 2/4 </primary>
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- If we knock of the $y^2$ term from the RHS we get for any solution,
- $\frac{d y}{d x}-x y \geq 0 .$

- Now we proceed exactly as we would with a linear differential equation and multiply by $\exp \left(-x^2 / 2\right)$.
- $\frac{d}{d x}\left(y(x) \exp \left(x^2 / 2\right)\right) \geq 0 .$
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<footer style = "font-size: 18px">Differential equations lecture-8 $\rightarrow$ Use of differential inequalities 1/4 $\rightarrow$ <span style = "color:blue"> Use of differential inequalities 2/4 </span> $\rightarrow$ Use of differential inequalities 3/4 $\rightarrow$ Use of differential inequalities 4/4 </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Use of differential inequalities 3/4 </primary>
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- We have a function $\phi(x) \geq 0$ on the interval $[a, b]$. Then we can say
- $\int_a^b \phi(x) d x \geq 0$

- Why? Since the integral is the area under the graph $y=\phi(x)$ bounded by the ordinates $x=a$ and $x=b$ and the $x-a x i s ~ y=0$. The inequality says that the graph is above the $x$ axis. The area under the graph is positive. In particular if we have two functions $f(x)$ and $g(x)$ and if $f(x) \geq g(x)$ on $[a, b]$, then
- $\int_a^b f(x) d x \geq \int_a^b g(x) d x .$

- How? Simply take $\phi(x)=f(x)-g(x)$ and we get the result.

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<footer style = "font-size: 18px">Use of differential inequalities 1/4 $\rightarrow$ Use of differential inequalities 2/4 $\rightarrow$ <span style = "color:blue"> Use of differential inequalities 3/4 </span> $\rightarrow$ Use of differential inequalities 4/4 $\rightarrow$ Exercise-1 </footer>

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### <primary style="font-size:24px"> Exercise-1 </primary>
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- Next, we show that $y(x) \equiv 0$ on some piece $[0, c]$ where $c>0$.
- Well, since $y(0)=0$ by continuity, $y(x)<1$ on say $[0, c]$. Thus
- $ (y(x))^2 \leq y(x)$

- Hence from (4.1) we get
- $\frac{d y}{d x} \leq(x+1) y, \quad 0 \leq x \leq c .$

- Exercise:
- Again, Imitate the idea of solving a linear differential equation and we get
- $ y(x) \leq 0, \quad 0 \leq x \leq c .$

- Can we conclude $y(x)=0$ on all of $x \geq 0$ ? Be careful !

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<footer style = "font-size: 18px">Use of differential inequalities 3/4 $\rightarrow$ Use of differential inequalities 4/4 $\rightarrow$ <span style = "color:blue"> Exercise-1 </span> $\rightarrow$ Examples on differential inequalities $\rightarrow$ Uniqueness theorem </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Examples on differential inequalities </primary>
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- Further examples on differential inequalities
- (1) How would you show that $y(x) \equiv 0$ is the only solution of
- $\frac{d y}{d x}-y\left(y \mid \sqrt{1}+x^3\right), \quad y(0)-0 ?$

- Try it now with
- $\frac{d y}{d x}-y\left(y^3+\sin ^2 x\right), \quad y(0)=0$

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<footer style = "font-size: 18px">Use of differential inequalities 4/4 $\rightarrow$ Exercise-1 $\rightarrow$ <span style = "color:blue"> Examples on differential inequalities </span> $\rightarrow$ Uniqueness theorem $\rightarrow$ Exercise-2 </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Uniqueness theorem </primary>
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- <ins>Towards a uniqueness theorem!</ins>
- What is the point of this exercise? Isn't there a simpler and more direct approach? Each time we encounter a situation such as the above where we could not divide by $y^2$ due to the fact that $y(0)=0$ are we to go through this rigmarole?

- We see the need for a general uniqueness theorem that could be directly employed instead of having to go through this rigmarole each time.

- Most Importantly. This simple example illustrates an important pattern of reasoning that can be used to prove the very general uniqueness theorem namely, by obtaining a linear differential inequality and then imitating the method for solving a linear differential equation!

- We shall not prove the general uniqueness theorem but the key ingredient in the proof is the excercise in the following slide.

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<footer style = "font-size: 18px">Exercise-1 $\rightarrow$ Examples on differential inequalities $\rightarrow$ <span style = "color:blue"> Uniqueness theorem </span> $\rightarrow$ Exercise-2 $\rightarrow$ Exercise-2 </footer>

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- <ins>Towards a uniqueness theorem contd...</ins>
- Exercise: Suppose $f(t)$ is a non-negative function on $[a, b]$ such that for certain POSITIVE constants $A$ and $B$,
- $ f(t) \leq A+B \int_a^t f(s) d s$
- then,
- $ f(t) \leq A \exp B(t-a)$

- Let us work out this exercise in some detail. Call the RHS of $(4.4)$ as $F(t)$ namely
- $ F(t)-A+B \int_a^t f(s) d s$
- and note that (4.4) says
- $ f(t) \leq F(t) . \quad F(a)=A .$

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<footer style = "font-size: 18px">Examples on differential inequalities $\rightarrow$ Uniqueness theorem $\rightarrow$ <span style = "color:blue"> Exercise-2 </span> $\rightarrow$ Exercise-2 $\rightarrow$ Energy equation </footer>

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### <primary style="font-size:24px"> Exercise-2 </primary>
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- Differentiating (4.6) we get the differential inequality:
- $\frac{d F}{d t}=B f(t) \leq B F(t)$

- If we multiply this through by $\exp (-B t)$, we see easily
- $\frac{d}{d t}(F(t) \exp (-B t))<0 .$

- Integrating over the interval $[a, t]$ and taking note of $F(a)=A$,
- $ F(t) \exp (-B(t-a))<A $
- which is (4.5).
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<footer style = "font-size: 18px">Uniqueness theorem $\rightarrow$ Exercise-2 $\rightarrow$ <span style = "color:blue"> Exercise-2 </span> $\rightarrow$ Energy equation $\rightarrow$ Energy equation contd... </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Energy equation </primary>
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- The energy equation

- Let us multiply the pendulum equation
- $\frac{d^2 y}{d t^2}+\frac{g}{l} \sin y=0$
- by the factor $\frac{d y}{d t}$ and the result can be expressed as
- $\frac{d}{d t}\left(\left(\frac{d y}{d t}\right)^2-\frac{2 g}{l} \cos y\right)=0$

- Integrating, we get the energy equation
- $\left(\frac{d y}{d t}\right)^2-\frac{2 g}{1} \cos y-E$

- Exercise: Identify the terms representing kinetic and potential energy perhaps after adding and subtracting a certain constant term

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<footer style = "font-size: 18px">Exercise-2 $\rightarrow$ Exercise-2 $\rightarrow$ <span style = "color:blue"> Energy equation </span> $\rightarrow$ Energy equation contd... $\rightarrow$ Elliptic integrals </footer>

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- Energy equation contd...
- Let us prescribe the initial conditions
- $ y(0)=0, \quad y^{\prime}(0)=c>0 .$

- The pendulum starts from the mean position with angular velocity c. Incorportating these into the energy equation we get
- $\left(\frac{dy}{dt}\right)^2=\frac{2 g}{1} \cos y-c^2-\frac{2 g}{l}$

- A more convenient form of this is
- $\left(\frac{d y}{d t}\right)^2=c^2-\frac{4 g}{1} \sin ^2\left(\frac{y}{2}\right)$

- It is tempting to take the square-root and sols by separating the variables. But you would run into elliptic integrals. Try it out or see A. I. Markushevich, Remarkable sine functions, American Elsevier Pub. Co, 1966.

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<footer style = "font-size: 18px">Exercise-2 $\rightarrow$ Energy equation $\rightarrow$ <span style = "color:blue"> Energy equation contd... </span> $\rightarrow$ Elliptic integrals $\rightarrow$ Elliptic integrals </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Elliptic integrals </primary>
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* Elliptic integrals enter the scene

- We have
- $\left(\begin{array}{l} d y \\ d t \end{array}\right)^2-c^2 \quad 4 g \sin ^2\left(\begin{array}{l} y \\- 2 \end{array}\right) $

- Since $\dot{y}(0)-c>0$, let us take the positive square-root and we get
- $\frac{d y}{d t}=\sqrt{c^2-\frac{4 g}{l} \sin ^2\left(\frac{y}{2}\right)}$

- We now integrate this variable separable equation over $[0, t]$ noting that $y(0)=0$ :
- $\int_0^{y(t)} \frac{d y}{\sqrt{c^2-\frac{4}{+} \sin ^2\left(\frac{z}{2}\right)}}=t .$

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<footer style = "font-size: 18px">Energy equation $\rightarrow$ Energy equation contd... $\rightarrow$ <span style = "color:blue"> Elliptic integrals </span> $\rightarrow$ Elliptic integrals $\rightarrow$ Pendulum and the elliptic functions </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Pendulum and the elliptic functions </primary>
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- Different types of motion
- We shall draw some qualitative conclusions from $(4$ 13) Note that if $c^2>4g I $ then $(4.13)$ shows that $\frac{d y}{d}$ is never zero Since initially it is positive, it must stay positive forevet and so $y(t)$ is monotone increasing with time.

- Since the RHS of (4.13) has least value $c^2 \quad 4 ;-a^2$ (say), we see that
- $\frac{d y}{d t} \geq a$
- for all values of $t$ and so $y(t)>2 t$. So the angle $y(t)$ keeps increasing unboundedly which means the pendulum must complete the whole circle performing a circular motion Note however that after completing one circuit the angle takes its value in $[2 \pi, 4 \pi \mid$ and thereafter takes its value in $[4 \pi, 6 \pi]$ etc.

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<footer style = "font-size: 18px">Elliptic integrals $\rightarrow$ Elliptic integrals $\rightarrow$ <span style = "color:blue"> Pendulum and the elliptic functions </span> $\rightarrow$ Different types of motion $\rightarrow$ Energy equation contd... </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
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- Different types of motion contd.

- Next, let us take up the case $c^2<\frac{4 \beta}{1}$
- rxercises:
- (2) Show that the angular velocity $\frac{d y}{d t}$ must become zero and $y(t)$ must attain a maximum in some finite time to. Obviously then we would call $4 t_0$ as the period of the pendulum. Suppose not Then $y(t)$ must strictly increase and it cannot reach \% otherwise RHS of (4.13) would become negative. Thus $y(t)$ must have a finite limit a as $t \rightarrow \infty$. But then (4.13) says that $y^{\prime}(t)$ mhist have a limit as $t \rightarrow \infty$. prove that if $f(t)$ is a differentiable function over $[0, \infty)$ such that both $f(t)$ and $f^{\prime}(t)$ have finite limits as $t \rightarrow \infty$ then $f^{\prime}(t) \longrightarrow 0$ as $t \rightarrow \infty$. Use Lagrange's Mean Value Theorem on an interval $[T, T+1]$ and also explain geometrically what is going on Now, to Complete the exercise,

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<footer style = "font-size: 18px">Elliptic integrals $\rightarrow$ Pendulum and the elliptic functions $\rightarrow$ <span style = "color:blue"> Different types of motion </span> $\rightarrow$ Energy equation contd... $\rightarrow$ Pendulum equation </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
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- Energy equation contd
- Let us prescribe the initial conditions
- $ y(0)=0, \quad y^{\prime}(0)-c>0 .$

- The pendulum starts from the mean position with angular velocity $c$ Incorportating these into the energy equation we get
- $\left(\begin{array}{l} d y \\ d t\end{array}\right)^2-\frac{2 g}{l} \cos y-c^2 \frac{2 g}{l} $

- A more convenient form of this is
- $\left(\frac{d y}{d t}\right)^2=c^2-\frac{4 g}{1} \sin ^2\left(\frac{y}{2}\right)$

- It is tempting to take the square-root and solve by separating the variables. But you would run into elliptic integrals Try it out or see A I Markushevich, Remarkable sine functions, American Elsevier Pub Co, 1966.

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<footer style = "font-size: 18px">Pendulum and the elliptic functions $\rightarrow$ Different types of motion $\rightarrow$ <span style = "color:blue"> Energy equation contd... </span> $\rightarrow$ Pendulum equation $\rightarrow$ Different types of motion contd... </footer>

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- The case $c^2<\frac{4 \pi}{1}$ contd...

- Now $\xi(t)$ monotone increases to a limit, equation $y^{\prime \prime}+\frac{8}{7} \sin y=0$ shows $y^{\prime \prime}(t)$ has a limit as $t \rightarrow \infty$

- But then $y^{\prime}(t)$ and $y^{\prime \prime}(t)$ have finite limits as $t \rightarrow \infty$ and by calculus (LMVT) argument.
- $\lim _{t \rightarrow \infty} y^{\prime \prime}(t)=0$

- Pendulum equation $y^{\prime \prime}+\frac{5}{7} \sin y=0$ would now force upon us
- $\lim _{t \rightarrow \infty} y(t)=0 .$

- In otherwords the maximum value of $y(t)$ is zero and so the pendulum is stationary! which is a contradiction.

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<footer style = "font-size: 18px">Different types of motion $\rightarrow$ Energy equation contd... $\rightarrow$ <span style = "color:blue"> Pendulum equation </span> $\rightarrow$ Different types of motion contd... $\rightarrow$ The amplitude </footer>

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- Different types of motion contd.
- (3) Explain why the point to obtained in the last exercise must be a local maximum.
- (4) What happens after time $t_0$ ? Show that $y(t)$ must attain a local a minimum value $\beta$
- (5) For example. let $z(t)--y(-t)$ and show that both $y(t)$ and $z(t)$ satisfy the pendulum equation and
- $ z(0)=y(0), \quad z^{\prime}(0)-y^{\prime}(0)$

- So by the uniqueness theorem (which we have not even stated!) we could conclude $z(t)=y(t)$ and hence the solution $y(t)$ must be an odd function.
- (6) Can you conclude that $\alpha=-\beta$ ?

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<footer style = "font-size: 18px">Energy equation contd... $\rightarrow$ Pendulum equation $\rightarrow$ <span style = "color:blue"> Different types of motion contd... </span> $\rightarrow$ The amplitude $\rightarrow$ Pendulum exhibits oscillatory motion </footer>

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- You can also do exercise (6) above by a different method. Assume that the maximum amplitude is $\alpha$ and the minimum amplitude is $-\beta$. Look at the eneray daruation
- $\left(\begin{array}{l} d y \\ d t\end{array}\right)^2=c^2 \quad 4 g \sin ^2\left(\begin{array}{l}- y \\ 2\end{array}\right)$

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<footer style = "font-size: 18px">Pendulum equation $\rightarrow$ Different types of motion contd... $\rightarrow$ <span style = "color:blue"> The amplitude </span> $\rightarrow$ Pendulum exhibits oscillatory motion $\rightarrow$ Different types of motion contd... </footer>

### <Success style="font-size:25px">Differential Equations L-8</Success>
### <primary style="font-size:24px"> Pendulum exhibits oscillatory motion </primary>
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- The LHS is zero when $y$ takes the values $\alpha \alpha-\beta$ and so equating the values on the RHS in the two cases:
- $\sin ^2\left(\frac{\alpha}{2}\right)=\sin ^2\left(\frac{\beta}{2}\right)$

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<footer style = "font-size: 18px">Different types of motion contd... $\rightarrow$ The amplitude $\rightarrow$ <span style = "color:blue"> Pendulum exhibits oscillatory motion </span> $\rightarrow$ Different types of motion contd... $\rightarrow$ Exercise </footer>

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- Different types of motion contd...The critical case
- Thus we see that if $c^2<4 g / I$ then the pendulum exhibits oscillatory motion and if $c^2>4 g / I$ then it exhibits circular motions. What happens if $c^2=\frac{4 \pi}{1}$ ? Is it that the pendulum takes forever to reach the point with $y(t)-\pi$ ?

- It turns out that the equation of the pendulum CAN be solved completely in the case $c^2-4 \beta$. This case is called the critical case

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<footer style = "font-size: 18px">The amplitude $\rightarrow$ Pendulum exhibits oscillatory motion $\rightarrow$ <span style = "color:blue"> Different types of motion contd... </span> $\rightarrow$ Exercise $\rightarrow$ Exercise </footer>

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- <ins> The critical case contd...</ins>

- The differential equation now reads:
- $\left(\frac{d y}{d t}\right)^2=\frac{4 g}{1} \cos ^2\left(\frac{y}{2}\right)$

- Now we observe that if $y^{\prime}\left(t_0\right)=0$ at some finite time $t_0$ then the RHS of the last equation would give
- $ y\left(t_0\right)=\pi, \quad y^{\prime}\left(t_0\right)=0$

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<footer style = "font-size: 18px">Pendulum exhibits oscillatory motion $\rightarrow$ Different types of motion contd... $\rightarrow$ <span style = "color:blue"> Exercise </span> $\rightarrow$ Exercise $\rightarrow$ Exercise </footer>

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- <ins>The critical case contd...</ins>

- We have the following variable separable equation for the critical case:
- $\frac{d y}{d t}-2 \sqrt{\frac{g}{l}} \cos (y / 2), \quad y(0)=0 .$

- Exercise:
- (7) Solve the equation (4.16) with the given initial condition.

- The answer is given by
- $\log (\sec (y / 2)+\tan (y / 2))-\sqrt{\frac{\pi}{1}} t$

- Show further that as $t, \infty, y(t) \quad>\pi$.
- The pendulum continuously ascends and as it does it loses its energy in such a manner that it takes infinitely long to reach the topmost pointl
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<footer style = "font-size: 18px">Exercise $\rightarrow$ Exercise $\rightarrow$ <span style = "color:blue"> Exercise </span> $\rightarrow$ Exercise $\rightarrow$ Exercise </footer>

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- <ins>Exercise</ins>:
- (8) Show that
- $\theta(s)=2 \tan ^{-1}\left(e^s\right)-\frac{\pi}{2}$

- Further show that the Gudermannian is an odd function and is increasing on R. (JEC2008, paper - 2).
- (9) Demonstrate the use of this to compute integrals. Show that
- $\int_0^{\pi / 3} \sec ^{1 /} \theta d \theta=\frac{1}{2^{16}} \int_0^{\log (2+\sqrt{3})}\left(e^s+e^{-s}\right)^{16} d s .$
- (10) Transform the integral (JEE 2014, paper - 2)
- $\int_{\pi / 4}^{\pi / 2}(2 \operatorname{cosec} \theta)^{17} d \theta$

- Hint: Put $\operatorname{cosec} \theta-\cot \theta-c^u$
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<footer style = "font-size: 18px">Exercise $\rightarrow$ Exercise $\rightarrow$ <span style = "color:blue"> Exercise </span> $\rightarrow$ Exercise $\rightarrow$ Thank you </footer>

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- <ins>Solution to problem 9</ins>
- To compute the integral
- $\int_0^{\pi / 3} \sec ^{17} \theta d \theta$

- Let us emplop the Gudermannian function $\theta(s)$ namely, put
- $\sec \theta+\tan \theta-c^3 \text {, hence } \sec \theta \quad \tan \theta-c^{-3}$
- and $\operatorname{sosec} \theta=\cosh$ s. Differentiating we get for the substitution theorem,

- $ \sec \theta \tan \theta d \theta  =\sinh s d s $
- $ \therefore \sec \theta d \theta  =(\tan \theta)^{-1} \sinh s d s $
- but $ \tan \theta  =\sinh s $
- so $ \sec \theta d \theta  =d s $
- $ \therefore \int(\sec \theta)^{17} d \theta-\int(\cosh s)^{16} d s $

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<footer style = "font-size: 18px">Exercise $\rightarrow$ Exercise $\rightarrow$ <span style = "color:blue"> Exercise </span> $\rightarrow$ Thank you $\rightarrow$  </footer>