Integral of 10/((1-1t^2)^0.5)

\int\frac{10}{\sqrt{1-t^2}}dt

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Answer

$10arcsin\left(t\right)+C_0$

Step by step solution

Problem

$\int\frac{10}{\sqrt{1-t^2}}dt$
1

Solve the integral $\int\frac{10}{\sqrt{1-t^2}}$ by trigonometric substitution using the substitution

$\begin{matrix}t=\sin\left(\theta\right) \\ dt=\cos\left(\theta\right)d\theta\end{matrix}$
2

Substituting in the original integral, we get

$\int\frac{10\cos\left(\theta\right)}{\sqrt{1-\sin\left(\theta\right)^2}}d\theta$
3

Applying the trigonometric identity: $1-\sin\left(\theta\right)^2=\cos\left(\theta\right)^2$

$\int\frac{10\cos\left(\theta\right)}{\sqrt{\cos\left(\theta\right)^2}}d\theta$
4

Applying the power of a power property

$\int\frac{10\cos\left(\theta\right)}{\cos\left(\theta\right)}d\theta$
5

Simplifying the fraction by $\cos\left(\theta\right)$

$\int10d\theta$
6

The integral of a constant is equal to the constant times the integral's variable

$10\theta$
7

Expressing the result of the integral in terms of the original variable

$10arcsin\left(t\right)$
8

Add the constant of integration

$10arcsin\left(t\right)+C_0$

Answer

$10arcsin\left(t\right)+C_0$

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Problem Analysis

Main topic:

Integration by trigonometric substitution

Time to solve it:

0.27 seconds

Views:

128