inte^x[((sin^-1x)sqrt(1-x^2)+1)/sqrt(1-x^2)]dx∫ex[(sin−1x)√1−x2+1√1−x2]dx ?
2 Answers
Explanation:
We have:
inte^x[((sin^-1x)sqrt(1-x^2)+1)/sqrt(1-x^2)]dx∫ex[(sin−1x)√1−x2+1√1−x2]dx
Splitting up the fraction, this becomes:
=inte^x[sin^-1x+1/sqrt(1-x^2)]dx=∫ex[sin−1x+1√1−x2]dx
Split up the integrals:
=inte^x(sin^-1x)dx+inte^x/sqrt(1-x^2)dx=∫ex(sin−1x)dx+∫ex√1−x2dx
Attempt to integrate
Let
Thus we see that the original integral equals:
=[e^x(sin^-1x)-inte^x/sqrt(1-x^2)dx]+inte^x/sqrt(1-x^2)dx=[ex(sin−1x)−∫ex√1−x2dx]+∫ex√1−x2dx
The integral
=e^x(sin^-1x)=ex(sin−1x)
Add the constant of integration:
=e^x(sin^-1x)+C=ex(sin−1x)+C
Explanation:
There is a pattern surrounding integrals involving
Since:
d/dxe^xf(x)=e^xf(x)+e^xf'(x)=e^x[f(x)+f'(x)]
Then:
inte^x[f(x)+f'(x)]dx=e^xf(x)+C
This is the case for the given problem, which simplifies to be:
inte^x[sin^-1x+1/sqrt(1-x^2)]dx
If
Thus the integral equals
=e^xf(x)+C=e^x(sin^-1x)+C
