MHT CET · Maths · Indefinite Integration
\(\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x=\)
- A \(\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c\)
- B \(\frac{2}{5}\left(\cos ^{-1} x\right)^{\frac{5}{2}}+c\)
- C \(\frac{5}{2}\left(\cos ^{-1} x\right)^{\frac{5}{2}}+c\)
- D \(\frac{5}{2}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x\)
Put \(\sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t\)
\(\begin{aligned} \therefore I &=\int t^{\frac{3}{2}} d t \\ &=\frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}=\frac{2}{5} t^{\frac{5}{2}}+c=\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c \end{aligned}\)
Put \(\sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t\)
\(\begin{aligned} \therefore I &=\int t^{\frac{3}{2}} d t \\ &=\frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}=\frac{2}{5} t^{\frac{5}{2}}+c=\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c \end{aligned}\)
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