A) \[A=\frac{1}{5},B=\frac{-2}{5\sqrt{15}},f(x)=\frac{4\tan x+3}{\sqrt{15}}\]
B) \[A=-\frac{1}{5},B=\frac{1}{\sqrt{15}},f(x)=\frac{4\tan x\left( \frac{x}{2} \right)+1}{\sqrt{15}}\]
C) \[A=\frac{2}{5},B=\frac{-2}{5},f(x)=\frac{4\tan x+1}{5}\]
D) \[A=\frac{2}{5},B=\frac{-2}{5\sqrt{15}},f(x)=\frac{4\tan \frac{x}{2}+1}{\sqrt{15}}\]
Correct Answer: D
Solution :
Let\[I=\int{\frac{1}{(\sin x+4)(\sin x-1)}}dx\] \[=\frac{1}{5}\int{\frac{(\sin x+4)-(\sin x-1)}{(\sin x+4)(\sin x-1)}}dx\] \[=\frac{1}{5}\int{\frac{2dt}{2t-1-{{t}^{2}}}}-\frac{1}{5}\int{\frac{2dt}{[2t+4(1+{{t}^{2}})]}}dx\] [putting\[\tan \frac{\alpha }{2}=t\]] \[\therefore \]\[I=-\frac{2}{5}\int{\frac{dt}{2t-1-{{t}^{2}}}-\frac{1}{10}\int{\frac{2dt}{{{t}^{2}}+\frac{1}{2}t+1}}}\] \[=\frac{-2}{5}\int{\frac{1}{{{(t-1)}^{2}}}}dt-\frac{1}{10}\int{\frac{1}{{{\left( 1+\frac{1}{4} \right)}^{2}}+{{\left( \frac{\sqrt{15}}{4} \right)}^{2}}}}\] \[=\frac{2}{5}\frac{1}{(t-1)}-\frac{2}{5\sqrt{15}}{{\tan }^{-1}}\left( \frac{4t+1}{\sqrt{15}} \right)+C\] \[=\frac{2}{5}\frac{1}{\tan \frac{x}{2}-1}=\frac{2}{5\sqrt{15}}{{\tan }^{-1}}\left( \frac{4\tan \frac{x}{2}+1}{\sqrt{15}} \right)+C\] \[\therefore \]\[A=\frac{2}{5},B=\frac{-2}{5\sqrt{15}},f(x)=\frac{4\tan \frac{x}{2}+1}{\sqrt{15}}\]
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