KCET · Maths · Indefinite Integration
\(\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x\) is equal to
- A \(\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C\)
- B \(\frac{1}{2} \cos ^{-1}\left(\sin ^{2} \mathrm{x}\right)+\mathrm{C}\)
- C \(\tan ^{-1}\left(\sin ^{2} \mathrm{x}\right)+\mathrm{C}\)
- D \(\tan ^{-1}\left(2 \sin ^{2} \mathrm{x}\right)+\mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x\)
Put \(\sin ^{2} \mathrm{x}=\mathrm{t} \Rightarrow 2 \sin \mathrm{x} \cos \mathrm{xdx}=\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{2 \sqrt{1-\mathrm{t}^{2}}}\)
\(=\frac{1}{2} \sin ^{-1} \mathrm{t}+\mathrm{C}\)
\(=\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C\)
Put \(\sin ^{2} \mathrm{x}=\mathrm{t} \Rightarrow 2 \sin \mathrm{x} \cos \mathrm{xdx}=\mathrm{dt}\)
\(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{2 \sqrt{1-\mathrm{t}^{2}}}\)
\(=\frac{1}{2} \sin ^{-1} \mathrm{t}+\mathrm{C}\)
\(=\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C\)
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