Example Question - solving integrals
Here are examples of questions we've helped users solve.
Solving an Arcsecant Integral
The integral in the image is: \[ \int_0^{\frac{1}{2}} \frac{dx}{\sqrt{1 - x^2}} \] This integral is the integral of the arcsecant derivative. We can use the substitution \( x = \sin(\theta) \) to solve it: \[ dx = \cos(\theta) d\theta \] When \( x = 0 \), \( \sin(\theta) = 0 \), so \( \theta = 0 \). When \( x = \frac{1}{2} \), \( \sin(\theta) = \frac{1}{2} \), which means \( \theta = \frac{\pi}{6} \) since it lies in the first quadrant. Substituting \( x = \sin(\theta) \) into the integral gives us: \[ \int_0^{\frac{\pi}{6}} \frac{\cos(\theta)}{\sqrt{1 - \sin^2(\theta)}} d\theta \] Now, \( \sqrt{1 - \sin^2(\theta)} = \sqrt{\cos^2(\theta)} = \cos(\theta) \) So the integral simplifies to: \[ \int_0^{\frac{\pi}{6}} d\theta \] This is just the integral of \(1\) with respect to \( \theta \) from \(0\) to \(\frac{\pi}{6}\), so it equals \( \theta \) evaluated between these two limits: \[ \theta \Big|_0^{\frac{\pi}{6}} = \frac{\pi}{6} - 0 = \frac{\pi}{6} \] Therefore, the integral equals \( \frac{\pi}{6} \).
Solving Integral of a Polynomial Function
The image shows an integral that you'd like to solve: \[ \int (8x^3 - x^2 + 5x - 1) \, dx \] To solve this integral, integrate each term individually with respect to \( x \). When integrating a polynomial term of the form \( x^n \), the antiderivative is \( \frac{x^{n+1}}{n+1} \) provided that \( n \neq -1 \). \[ \int 8x^3 \, dx = \frac{8}{4}x^{4} = 2x^4 \] \[ \int (-x^2) \, dx = -\frac{1}{3}x^{3} \] \[ \int 5x \, dx = \frac{5}{2}x^{2} \] \[ \int (-1) \, dx = -x \] Now summing all of the individual antiderivatives and adding a constant of integration \( C \), we have: \[ 2x^4 - \frac{1}{3}x^{3} + \frac{5}{2}x^{2} - x + C \] So the integral of \( 8x^3 - x^2 + 5x - 1 \) with respect to \( x \) is: \[ \int (8x^3 - x^2 + 5x - 1) \, dx = 2x^4 - \frac{1}{3}x^{3} + \frac{5}{2}x^{2} - x + C \]
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