Trig Substitutions — Question 2

Question 2 – Solution

The radical has the form $\sqrt{a^2-x^2}$, so let $$x=5\sin\theta,\qquad dx=5\cos\theta\,d\theta.$$ Choose $-\pi/2<\theta<\pi/2$, so $\cos\theta>0$. Then $$\sqrt{25-x^2}=\sqrt{25-25\sin^2\theta}=5\cos\theta.$$ Substitution gives $$\begin{align*} I&=\int\frac{25\sin^2\theta}{5\cos\theta} (5\cos\theta\,d\theta)\\ &=25\int\sin^2\theta\,d\theta\\ &=\frac{25}{2}\int(1-\cos2\theta)\,d\theta\\ &=\frac{25}{2}\theta-\frac{25}{4}\sin2\theta+C. \end{align*}$$ Since $\theta=\arcsin(x/5)$ and $$\sin2\theta=2\sin\theta\cos\theta =2\left(\frac{x}{5}\right)\left(\frac{\sqrt{25-x^2}}5\right),$$ we obtain $$\boxed{\displaystyle I=\frac{25}{2}\arcsin\left(\frac{x}{5}\right) -\frac{x}{2}\sqrt{25-x^2}+C}.$$