Integral Calculus
86359
If \(f(x)=\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) and \(g(x)=\mathrm{e}^{\sin ^{-1} x}\), then \(\int f(x) g(x)\) is equal to
1 \(\mathrm{e}^{\sin ^{-1} x}\left(\sin ^{-1} x-1\right)+C\)
2 \(\mathrm{e}^{\sin ^{-1} x}+\mathrm{C}\)
3 \(\mathrm{e}^{\left(\sin ^{-1} x\right)^{2}}+\mathrm{C}\)
4 \(\mathrm{e}^{2 \sin ^{-1} x}+C\)
5 \(\mathrm{e}^{\sin ^{-1} x} \sin ^{-1} x+C\)
Explanation:
(A) : \(I=\int f(x) g(x) d x=\int \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \cdot e^{\sin ^{-1}} x d x\)
Let, \(\quad \sin ^{-1} \mathrm{x}=\mathrm{t}\)
\(\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\mathrm{dt}\)
Now, \(\quad I=\int t . e^{t} d t\)
\(I=t^{t}-\int 1 . e^{t} d t\)
\(I=t^{t}-e^{t}+C\)
\(\mathrm{I}=\mathrm{e}^{\mathrm{t}}(\mathrm{t}-1)+\mathrm{C}\)
\(I=\mathrm{e}^{\sin ^{-1} x} \mathrm{x}\left(\sin ^{-1} \mathrm{x}-1\right)+\mathrm{C} \quad\left(\because \mathrm{t}=\sin ^{-1} \mathrm{x}\right)\)