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\(\smallint \frac{{{\rm{sin}}\frac{{5{\rm{x}}}}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\)is equal to(where c is a constant of integration.)1. 2x + sin x + 2 sin 2x + C2. x + 2 sin x + 2 sin 2x + C3. x + 2 sin x + sin 2x + C4. 2x + sin x – sin 2x + C |
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Answer» Correct Answer - Option 3 : x + 2 sin x + sin 2x + C \(I = \smallint \frac{{{\rm{sin}}\left( {\frac{{5{\rm{x}}}}{2}} \right)}}{{{\rm{sin}}\left( {\frac{{\rm{x}}}{2}} \right)}}{\rm{dx}}\) \(= \smallint \frac{{{\rm{sin}}\left( {2x + \frac{x}{2}} \right)}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\) \(= \smallint \frac{{\sin 2x\cos \frac{x}{2} + \cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\) \(= \smallint \frac{{2\sin x\cos x\cos \frac{x}{2} + \cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\) [∵ sin 2x = 2 sin x cos x] \(= \smallint \frac{{2\sin x\cos x\cos \frac{x}{2} + \cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\) \(= \smallint \frac{{2 \times 2\sin \frac{x}{2}\cos \frac{x}{2}\cos x\cos \frac{x}{2} + \cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\) \(\left[ \because{\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2}} \right]\) \(= \smallint \frac{{4\sin \frac{x}{2}\cos \frac{x}{2}\cos x\cos \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}} + \frac{{\cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\) \(= \smallint \left[ {4\cos x{{\cos }^2}\frac{x}{2} + \cos 2x} \right]{\rm{dx}}\) \(= \smallint \left[ {4\cos x\left( {\frac{{1 + \cos x}}{2}} \right) + \cos 2x} \right]{\rm{dx}}\) \(\left[ \because{\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}} } \right]\) = ∫[2 cos x (1 + cos x) + cos 2x] dx = ∫[2 cos x + 2 cos2 x + cos 2x] dx \(= \smallint \left[ {2\cos x + 2\left( {\frac{{1 + \cos 2x}}{2}} \right) + \cos 2x} \right]{\rm{dx}}\) = ∫[2 cos x + 1 + cos 2x + cos 2x] dx = ∫[2 cos x + 1 + 2 cos 2x] dx \(= 2{\rm{sinx}} + {\rm{x}} + \frac{{2{\rm{sin\;}}2{\rm{x}}}}{2} + {\rm{c}}\) ⇒ x + 2 sin x + sin 2x + c |
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