Question

4·49-25-

Answer 1

The given function is 1 9−25 x 2 .

Consider, ∫ 1 9−25 x 2 dx(1)

Also, ∫ 1 a 2 − x 2 = 1 a sin −1 x a +c (2)

Now let, 5 x =t 5dx=dt

Substitute values of t and dt in (1),

∫ 1 9− 25 2 dx= 1 5 ∫ ( 1 9− t 2 ) dt = 1 5 ∫ ( 1 3 2 − t 2 ) dt = 1 5 sin −1 ( t 3 )+c = 1 5 sin −1 ( 5x 3 )+c By Using (2)

Thus, the integral of the function 1 9−25 x 2 is 1 5 sin −1 ( 5x 3 )+c.