+0

Calculus..........

0
707
4

Feb 14, 2016

#3
+23278
+10

Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1.

$$\small{ \begin{array}{lrcll} &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } \\ \hline \text{Substitution : } & x &=& \sin{(u)} \qquad \rightarrow \qquad u = \arcsin{(x)}\\ & \ dx &=& \cos{(u)}\ du \\ \hline &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \qquad | \qquad 1-\sin^2{(u)} = \cos^2{(u)}\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{\cos^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \qquad | \qquad dx = \cos{(u)}\ du\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} }\cdot \cos{(u)}\ du } \\ & &=&\int \limits_{-1}^{1} { \ du } \\ & &=& [ u ]_{-1}^{1} \qquad | \qquad u = \arcsin{(x)} \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } & \mathbf{=} & \mathbf{ [~ \arcsin{(x)} ~]_{-1}^{1} }\\ \hline & \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } & = & [~ \arcsin{(x)} ~]_{-1}^{1} \\ & & = & [~ \arcsin{(1)}-\arcsin{(-1)} ~]\\ & & = & [~ \frac{\pi}{2} -\frac{-\pi}{2} ~]\\ & & = & [~ \frac{\pi}{2} +\frac{\pi}{2} ~]\\ & & = & \pi \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ \pi } \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ 3.14159265359 } \end{array} }$$

.
Feb 15, 2016

#1
+5

Compute the definite integral:

integral_(-1)^1 1/sqrt(1-x^2) dx

1/sqrt(1-x^2) has discontinuities at x = -1 and x = 1, which both produce improper bounds:

= integral_(-1)^1 1/sqrt(1-x^2) dx

Since 1/sqrt(1-x^2) is an even function and the interval [-1, 1] is symmetric about 0, integral_(-1)^1 1/sqrt(1-x^2) dx = 2 integral_0^1 1/sqrt(1-x^2) dx. Applying this identity will reduce the number of improper endpoints on the integration domain:

= 2 integral_0^1 1/sqrt(1-x^2) dx

Apply the fundamental theorem of calculus.

The antiderivative of 1/sqrt(1-x^2) is sin^(-1)(x):

= lim_(b->1^-) 2 sin^(-1)(x)|_0^b

Evaluate the antiderivative at the limits and subtract.

lim_(b->1^-) 2 sin^(-1)(x)|_0^b = (lim_(b->1^-) 2 sin^(-1)(b))-2 sin^(-1)(0) = (lim_(b->1^-) 2 sin^(-1)(b))-0:

= (lim_(b->1^-) 2 sin^(-1)(b))

lim_(b->1^-) 2 sin^(-1)(b) = pi:

Feb 14, 2016
#2
+104712
0
.
Feb 14, 2016
edited by CPhill  Feb 14, 2016
#3
+23278
+10

Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1.

$$\small{ \begin{array}{lrcll} &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } \\ \hline \text{Substitution : } & x &=& \sin{(u)} \qquad \rightarrow \qquad u = \arcsin{(x)}\\ & \ dx &=& \cos{(u)}\ du \\ \hline &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \qquad | \qquad 1-\sin^2{(u)} = \cos^2{(u)}\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{\cos^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \qquad | \qquad dx = \cos{(u)}\ du\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} }\cdot \cos{(u)}\ du } \\ & &=&\int \limits_{-1}^{1} { \ du } \\ & &=& [ u ]_{-1}^{1} \qquad | \qquad u = \arcsin{(x)} \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } & \mathbf{=} & \mathbf{ [~ \arcsin{(x)} ~]_{-1}^{1} }\\ \hline & \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } & = & [~ \arcsin{(x)} ~]_{-1}^{1} \\ & & = & [~ \arcsin{(1)}-\arcsin{(-1)} ~]\\ & & = & [~ \frac{\pi}{2} -\frac{-\pi}{2} ~]\\ & & = & [~ \frac{\pi}{2} +\frac{\pi}{2} ~]\\ & & = & \pi \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ \pi } \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ 3.14159265359 } \end{array} }$$

heureka Feb 15, 2016
#4
+105509
+5

My answer is the same as Heureka's, I just like to start it a little different.

$$\int_{-1}^1 \frac{1}{\sqrt{1 - x^2}}\;dx$$

I like to start with this little triangle  it wasn't meant to be that big.

$$cos\theta=\sqrt{1-x^2}\\~\\ sin\theta=x\\ x=\sin\theta\\ \frac{dx}{d\theta}=cos\theta\\ dx=cos\theta\;d\theta\\~\\ When\;\;x=1,\;\;\;\;sin\theta =1\;\;\;\;\;\;\;\;\rightarrow\;\;\;\theta=\frac{\pi}{2}\\ When\;\;x=-1,\;\;\;\;sin\theta =-1\;\;\;\;\rightarrow\;\;\;\theta=\frac{-\pi}{2}\\ SO\\$$

$$\displaystyle \int_{-1}^1 \frac{1}{\sqrt{1 - x^2}}\;dx\\~\\ =\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\;\;\; \frac{1}{cos\theta}\;cos\theta\;d\theta\\~\\ =\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\;\;\; 1\;d\theta\\~\\ =\left[\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\~\\ =\frac{\pi}{2}-\;-\frac{\pi}{2}\\~\\ =\pi$$

.
Feb 17, 2016