Find out the answer to "∫ex[(sin−1x)1−x2+11−x2]dx?"

jisu61hbke

Answered question

2022-03-29

\int e^{x} [\frac{(\sin^{- 1} x) \sqrt{1 - x^{2}} + 1}{\sqrt{1 - x^{2}}}] d x

crazyrocketrz5z

Beginner2022-03-30Added 11 answers

We have:

\int e^{x} [\frac{(\sin^{- 1} x) \sqrt{1 - x^{2}} + 1}{\sqrt{1 - x^{2}}}] d x

Splitting up the fraction, this becomes:

= \int e^{x} [\sin^{- 1} x + \frac{1}{\sqrt{1 - x^{2}}}] d x

Split up the integrals:

= \int e^{x} (\sin^{- 1} x) d x + \int \frac{e^{x}}{\sqrt{1 - x^{2}}} d x

Attempt to integrate

\int e^{x} (\sin^{- 1} x) d x

via integration by parts, which takes the form

\int u v - u v - \int v d u

Let

u = \sin^{- 1} x

. so

d u = \frac{1}{\sqrt{1 - x^{2}}} d x

, and

d v = e^{x} d x

, so

v = e^{x}

Thus we see that the original integral equals:

= [e^{x} (\sin^{- 1} x) - \int \frac{e^{x}}{\sqrt{1 - x^{2}}} d x] + \int \frac{e^{x}}{\sqrt{1 - x^{2}}} d x

The integral

\int \frac{e^{x}}{\sqrt{1 - x^{2}}} d x

will cancel with itself, leaving just

= e^{x} (\sin^{- 1} x)

Add the constant of integration:

= e^{x} (\sin^{- 1} x) + C

LiaisyAciskriuu

Beginner2022-03-31Added 10 answers

There is a pattern surrounding integrals involving

e^{x}

, in cases like this problem (and all the others surrounding it on the page).
Since:

\frac{d}{d x} e^{x} f (x) = e^{x} f (x) + e^{x} f^{'} (x) = e^{x} [f (x) + f^{'} (x)]

Then:

\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C

This is the case for the given problem, which simplifies to be:

\int e^{x} [\sin^{- 1} x + \frac{1}{\sqrt{1 - x^{2}}}] d x

f (x) = \sin^{- 1} x

, then

f^{'} (x) = \frac{1}{\sqrt{1 - x^{2}}}

Thus the integral equals

= e^{x} f (x) + C = e^{x} (\sin^{- 1} x) + C

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