MHT CET · Maths · Indefinite Integration
The value of \(\int \sqrt{1+\sec x} d x\) is
- A \(\sin ^{-1}(\sqrt{2} \sin x)+C\)
- B \(2 \sin ^{-1}(\sqrt{2} \sin x / 2)+C\)
- C \(2 \sin ^{-1}(\sqrt{2} \sin x)+C\)
- D \(2 \sin ^{-1}(\sqrt{2} x / 2)+C\)
Answer & Solution
Correct Answer
(B) \(2 \sin ^{-1}(\sqrt{2} \sin x / 2)+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \sqrt{1+\sec x} d x\)
\(=\int \frac{\sqrt{1+\cos x}}{\sqrt{\cos x}} d x\)
\(=\int \frac{\sqrt{2} \cdot \cos \frac{x}{2}}{\sqrt{1-2 \sin ^{2} \frac{x}{2}}} d x\)
(let \(\left.t=\sqrt{2} \sin \frac{x}{2} \Rightarrow d t=\frac{\sqrt{2}}{2} \cdot \cos \frac{x}{2} \cdot d x\right)\)
\(=\int \frac{2 d t}{\sqrt{1-t^{2}}}\)
\(=2 \sin ^{-1}(t)+C\)
\(=2 \sin ^{-1}\left(\sqrt{2} \cdot \sin \frac{x}{2}\right)+C\)
\(=\int \frac{\sqrt{1+\cos x}}{\sqrt{\cos x}} d x\)
\(=\int \frac{\sqrt{2} \cdot \cos \frac{x}{2}}{\sqrt{1-2 \sin ^{2} \frac{x}{2}}} d x\)
(let \(\left.t=\sqrt{2} \sin \frac{x}{2} \Rightarrow d t=\frac{\sqrt{2}}{2} \cdot \cos \frac{x}{2} \cdot d x\right)\)
\(=\int \frac{2 d t}{\sqrt{1-t^{2}}}\)
\(=2 \sin ^{-1}(t)+C\)
\(=2 \sin ^{-1}\left(\sqrt{2} \cdot \sin \frac{x}{2}\right)+C\)
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