$\int \frac{t^{2}}{\sqrt{36-t^{2}}} d t=\frac{1}{2} t \sqrt{36-t^{2}}+\frac{36}{2} \arcsin \frac{t}{6}+C$ .

Name: $\int \frac{t^{2}}{\sqrt{36-t^{2}}} d t=\frac{1}{2} t \sqrt{36-t^{2}}+\frac{36}{2} \arcsin \frac{t}{6}+C$ .
Brand: Quizplus
Rating: 5.0 (38 reviews)

Question

Quizplus · Accepted Answer

The Answer of $\int \frac{t^{2}}{\sqrt{36-t^{2}}} d t=\frac{1}{2} t \sqrt{36-t^{2}}+\frac{36}{2} \arcsin \frac{t}{6}+C$ .

∫t236−t2dt=12t36−t2+362arcsin⁡t6+C\int \frac{t^{2}}{\sqrt{36-t^{2}}} d t=\frac{1}{2} t \sqrt{36-t^{2}}+\frac{36}{2} \arcsin \frac{t}{6}+C∫36−t2​t2​dt=21​t36−t2​+236​arcsin6t​+C

$\int \frac{t^{2}}{\sqrt{36-t^{2}}} d t=\frac{1}{2} t \sqrt{36-t^{2}}+\frac{36}{2} \arcsin \frac{t}{6}+C$