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Trigonometry Calculator

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Example
Created on 2024-06-20Asked by Scarlett Harris (Solvelet student)
Evaluate the integral ∫1−x2 dx\int \sqrt{1 - x^2} \, dx using trigonometric substitution.

Solution

To evaluate the integral ∫1−x2 dx\int \sqrt{1 - x^2} \, dx using trigonometric substitution: 1. **Substitute x=sin⁡(θ)x = \sin(\theta):** dx=cos⁡(θ) dθ. dx = \cos(\theta) \, d\theta. 2. **Rewrite the integral:** ∫1−sin⁡2(θ)cos⁡(θ) dθ. \int \sqrt{1 - \sin^2(\theta)} \cos(\theta) \, d\theta. 3. **Simplify using cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta):** ∫cos⁡2(θ) dθ. \int \cos^2(\theta) \, d\theta. 4. **Use the identity cos⁡2(θ)=1+cos⁡(2θ)2\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}:** ∫1+cos⁡(2θ)2 dθ=12∫(1+cos⁡(2θ)) dθ. \int \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int (1 + \cos(2\theta)) \, d\theta. 5. **Integrate:** 12(θ+12sin⁡(2θ))+C. \frac{1}{2} \left( \theta + \frac{1}{2} \sin(2\theta) \right) + C. 6. **Substitute back x=sin⁡(θ)x = \sin(\theta):** θ=sin⁡−1(x). \theta = \sin^{-1}(x). sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)=2x1−x2. \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2x \sqrt{1 - x^2}. 7. **Result:** The integral ∫1−x2 dx\int \sqrt{1 - x^2} \, dx is: 12sin⁡−1(x)+14x1−x2+C. \frac{1}{2} \sin^{-1}(x) + \frac{1}{4} x \sqrt{1 - x^2} + C. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Mia Allen on Solvelet
1. Determine the length of side cc in the triangle below using the law of sines. [Diagram: Triangle ABC with angle C and sides aa, bb, and cc]2. Find the value of tan⁡(π4) \tan\left(\frac{\pi}{4}\right) .,
DefinitionTrigonometry is a part of mathematics that studies the angle, the length and distance between the points, because it can be a supplementary study of the geometricalexceptionapart studying in two dimensions as shown in the figure in C (I. i), and how it is related to the sine, thecosine, and the tangent of an angle. It includes sines-cosines, tangents things. Example: For a right triangle with hypotenuse c and angle θ, sin(θ)=copposite​, cos(θ)=cadjacent ​.
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