Question

under root a square minus x square ka samakalan​

Answer 1

Answer:

which class it is dear friend

Answer 2

hey mate

Root of a^2-x^2

The integral of a2−x2−−−−−−√ is of the form

I=∫a2−x2−−−−−−√dx=xa2−x2−−−−−−√2+a22sin−1(xa)+c

This integral can be written as

I=∫a2−x2−−−−−−√⋅1dx

Here the first function is a2−x2−−−−−−√ and the second function is 1

I=∫a2−x2−−−−−−√⋅1dx - - - (i)

Using the formula for integration by parts, we have

∫[f(x)g(x)]dx=f(x)∫g(x)dx−∫[ddxf(x)∫g(x)dx]dx

Using the formula above, equation (i) becomes

I=a2−x2−−−−−−√∫1dx−∫[ddxa2−x2−−−−−−√(∫1dx)]dx⇒I=xa2−x2−−−−−−√−∫−x2a2−x2−−−−−−√dx⇒I=xa2−x2−−−−−−√−∫−a2+a2−x2a2−x2−−−−−−√dx⇒I=xa2−x2−−−−−−√−∫−a2a2−x2−−−−−−√dx−∫a2−x2a2−x2−−−−−−√dx⇒I=xa2−x2−−−−−−√+a2∫1a2−x2−−−−−−√dx−∫a2−x2−−−−−−√dx⇒I=xa2−x2−−−−−−√+a2sin−1(xa)−I+c⇒I+I=xa2−x2−−−−−−√+a2sin−1(xa)+c⇒2I=xa2−x2−−−−−−√+a2sin−1(xa)+c⇒I=xa2−x2−−−−−−√2+a22sin−1(xa)+c⇒∫a2−x2−−−−−−√dx=xa2−x2−−−−−−√2