Question

The value of U% Va2 + x2 ) dx is​

Answer 1

Step-by-step explanation:

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+a22sin–1(xa)+c

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