A) \[\frac{1}{2}\ln \left| \frac{2+\sin x-\cos x}{2-\sin x+\cos x} \right|-\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sin x+\cos x}{\sqrt{2}} \right)+c\]
B) \[\frac{1}{2}\ln \left| \frac{2+\sin x-\cos x}{2-\sin x+\cos x} \right|-\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sin x+\cos x}{\sqrt{2}} \right)+c\]
C) \[\frac{1}{4}\ln \left| \frac{2+\sin x-\cos x}{2-\sin x+\cos x} \right|-\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sin x+\cos x}{\sqrt{2}} \right)+c\]
D) none of these
Correct Answer: C
Solution :
[c] \[I=\int{\frac{2\sin x}{(3+sin2x)}dx}\] \[=\int{\frac{\sin x+\cos x+\sin x-\cos x}{(3+sin2x)}dx}\] \[=\int{\frac{\sin x+\cos x}{3+\underset{\begin{smallmatrix} \downarrow \\ {{I}_{1}} \end{smallmatrix}}{\mathop{\sin }}\,2x}}dx-\int{\frac{-\sin x+\cos x}{(3+\underset{\begin{smallmatrix} \downarrow \\ {{I}_{2}} \end{smallmatrix}}{\mathop{\sin }}\,2x)}dx}\] Putting \[{{t}_{1}}=\sin x-\cos x\] in \[{{I}_{1}}\] and \[{{t}_{2}}=\sin x+\cos x\]in \[{{I}_{2}}\], we get \[I=\int{\frac{d{{t}_{1}}}{[3+(1-{{t}^{2}}_{1})]}-\int{\frac{d{{t}_{2}}}{[3+({{t}_{2}}^{2}-1)]}}}\] =\[\int{\frac{dt}{4-{{t}_{1}}^{2}}-\int{\frac{d{{t}_{2}}}{2+{{t}_{2}}^{2}}}}\] \[=\frac{1}{4}\ln \left| \frac{2+{{t}_{1}}}{2-{{t}_{1}}} \right|-\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{t}_{2}}}{\sqrt{2}} \right)+C\] \[=\frac{1}{4}\ln \left| \frac{2+\sin x-\cos x}{2-\sin x+\cos x} \right|-\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sin x+\cos x}{\sqrt{2}} \right)+C\]
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