# Thread: [SOLVED] integration using trigonometry substitution

1. ## [SOLVED] integration using trigonometry substitution

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{x^2} dx$ [ using the substitution, x = costheta]

$\displaystyle \frac{dx}{d\theta }= -sin\theta$

$\displaystyle dx = -sin\theta d\theta$

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-cos^2\theta }}{cos^2\theta } dx$

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-cos^2\theta }}{cos^2\theta } .-sin\theta$ dtheta

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{sin\theta }{cos^2\theta }.-sin\theta$ dtheta

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{- sin^2\theta}{cos^2\theta }$

never mind i got it

2. Hello differentiate
Originally Posted by differentiate
$\displaystyle \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-x^2}}{x^2} dx$ [ using the substitution, x = costheta]

$\displaystyle \frac{dx}{d\theta }= -sin\theta$

$\displaystyle dx = -sin\theta d\theta$

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-cos^2\theta }}{cos^2\theta } dx$

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{\sqrt{1-cos^2\theta }}{cos^2\theta } .-sin\theta$ dtheta

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{sin\theta }{cos^2\theta }.-sin\theta$ dtheta

$\displaystyle \int_{\frac{1}{2}}^{1}\frac{- sin^2\theta}{cos^2\theta }$

x = cos@
when x = 1,
@ = 0
when x = 1/2
@ = pi/3

what do I do next?
The integral becomes $\displaystyle \int_{0}^{\frac{\pi}{3}}\tan^2\theta d\theta$, which you can solve using $\displaystyle \tan^2\theta = \sec^2\theta-1$, and noting that $\displaystyle \int\sec^2\theta d\theta = \tan\theta$.