Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1. Please help and thanks.
+26387 Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1.
\(\small{ \begin{array}{lrcll} &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } \\ \hline \text{Substitution : } & x &=& \sin{(u)} \qquad \rightarrow \qquad u = \arcsin{(x)}\\ & \ dx &=& \cos{(u)}\ du \\ \hline &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \qquad | \qquad 1-\sin^2{(u)} = \cos^2{(u)}\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{\cos^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \qquad | \qquad dx = \cos{(u)}\ du\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} }\cdot \cos{(u)}\ du } \\ & &=&\int \limits_{-1}^{1} { \ du } \\ & &=& [ u ]_{-1}^{1} \qquad | \qquad u = \arcsin{(x)} \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } & \mathbf{=} & \mathbf{ [~ \arcsin{(x)} ~]_{-1}^{1} }\\ \hline & \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } & = & [~ \arcsin{(x)} ~]_{-1}^{1} \\ & & = & [~ \arcsin{(1)}-\arcsin{(-1)} ~]\\ & & = & [~ \frac{\pi}{2} -\frac{-\pi}{2} ~]\\ & & = & [~ \frac{\pi}{2} +\frac{\pi}{2} ~]\\ & & = & \pi \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ \pi } \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ 3.14159265359 } \end{array} }\)
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:
Answer: |= pi
+26387 Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1.
\(\small{ \begin{array}{lrcll} &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } \\ \hline \text{Substitution : } & x &=& \sin{(u)} \qquad \rightarrow \qquad u = \arcsin{(x)}\\ & \ dx &=& \cos{(u)}\ du \\ \hline &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \qquad | \qquad 1-\sin^2{(u)} = \cos^2{(u)}\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{\cos^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \qquad | \qquad dx = \cos{(u)}\ du\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} }\cdot \cos{(u)}\ du } \\ & &=&\int \limits_{-1}^{1} { \ du } \\ & &=& [ u ]_{-1}^{1} \qquad | \qquad u = \arcsin{(x)} \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } & \mathbf{=} & \mathbf{ [~ \arcsin{(x)} ~]_{-1}^{1} }\\ \hline & \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } & = & [~ \arcsin{(x)} ~]_{-1}^{1} \\ & & = & [~ \arcsin{(1)}-\arcsin{(-1)} ~]\\ & & = & [~ \frac{\pi}{2} -\frac{-\pi}{2} ~]\\ & & = & [~ \frac{\pi}{2} +\frac{\pi}{2} ~]\\ & & = & \pi \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ \pi } \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ 3.14159265359 } \end{array} }\)
+118667 Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1. Please help and thanks.
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. 
\(cos\theta=\sqrt{1-x^2}\\~\\ sin\theta=x\\ x=\sin\theta\\ \frac{dx}{d\theta}=cos\theta\\ dx=cos\theta\;d\theta\\~\\ When\;\;x=1,\;\;\;\;sin\theta =1\;\;\;\;\;\;\;\;\rightarrow\;\;\;\theta=\frac{\pi}{2}\\ When\;\;x=-1,\;\;\;\;sin\theta =-1\;\;\;\;\rightarrow\;\;\;\theta=\frac{-\pi}{2}\\ SO\\ \)
\(\displaystyle \int_{-1}^1 \frac{1}{\sqrt{1 - x^2}}\;dx\\~\\ =\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\;\;\; \frac{1}{cos\theta}\;cos\theta\;d\theta\\~\\ =\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\;\;\; 1\;d\theta\\~\\ =\left[\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\~\\ =\frac{\pi}{2}-\;-\frac{\pi}{2}\\~\\ =\pi\)
