Answer
$\dfrac{25}{2} \sin^{-1}(t/5)+\dfrac{t\sqrt {25 -t^2}}{2} +C$
Work Step by Step
Let us consider $t =5 \sin \theta $
and $dt=5 \cos \theta d \theta $
Now, the given integral becomes:
$\int \sqrt{25-25 \sin^2 \theta } (5 \cos \theta d \theta ) =\int \sqrt{(25(1-\sin^2 \theta) } (5 \cos \theta d \theta ) $
We know that $\cos^2 \theta=1-\sin^2 \theta$ and $\cos^2 \theta =(1/2)+(1/2) \cos 2 \theta$
Then, we have $\int 25 \cos^2 \theta = 25\int (\dfrac{1}{2}+\dfrac{1}{2} \cos ( 2\theta ))d \theta$
Now, integrate and plug in $ \sin \theta =t/5 \implies \theta =\sin^{-1}(t/5)$
Thus, $ 25 (\dfrac{1}{2} \theta+\dfrac{1}{4} \sin ( 2\theta ))d \theta +C=\dfrac{25}{2} \theta+\dfrac{25}{4} (2 \sin \theta \cos \theta) d \theta +C$
or, $=\dfrac{25}{2} \sin^{-1}(t/5)+\dfrac{t\sqrt {25 -t^2}}{2} +C$
