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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \large\frac{1}{5-4 \sin x} $$dx$

$(a)\;\frac{1}{2 \sqrt {41}} \log \bigg| \frac{t-4/5 - \sqrt {41}/5}{t- 4/5 +\sqrt {41}/5}\bigg|+c \\(b)\;\frac{1}{2 \sqrt {41}} \log \bigg| \frac{t+4/5 - \sqrt {41}/5}{t- 4/5 +\sqrt {41}/5}\bigg|+c \\(c)\; \frac{1}{\sqrt {41}} \log \bigg| \frac{t+4/5 - \sqrt {41}/5}{t- 4/5 +\sqrt {41}/5}\bigg|+c \\ (d)\;\frac{1}{ \sqrt {41}} \log \bigg| \frac{t-4/5 - \sqrt {41}/5}{t- 4/5 +\sqrt {41}/5}\bigg|+c$

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1 Answer

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$\int \large\frac{1}{5-4 \sin x} $$dx$
=> $\int \large\frac{1}{5-\Large\frac{8 \tan x/2}{1+ \tan ^2 x/2}} $$dx$
=> $\int \large\frac{\sec^2 x/2}{5-5 \tan ^2 x/2 + 8 \tan x/2}$$dx$
=>$\tan x/2 =t$
=>$\sec^2 x/2 \times \large\frac{1}{2} $$dx=2dt$
=>$\sec^2 \large\frac{x}{2} $$dx=2dt$
=> $\int \large\frac{2 dt}{5 -5t^2 +8t}$
=> $\large\frac{2}{5} $$\int \large\frac{dt}{1-t^2 +\Large\frac{8}{5}+\frac{16}{25}-\frac{16}{25}}$
=> $\large\frac{2}{5} \int \frac{dt}{1+\Large\frac{16}{25}- (t^2 -\Large\frac{8}{5}t+\frac{16}{25})}$
=> $\large\frac{2}{5} \int \frac{dt}{\bigg(\Large\frac{\sqrt {41}}{5}\bigg)^2- \bigg(t -\Large\frac{4}{5}\bigg)^2}$
$\large \frac{2}{5} \times \frac{1}{2 \times \Large\frac{\sqrt {41}}{5}} $$ \log \bigg| \frac{t-4/5 - \sqrt {41}/5}{t- 4/5 +\sqrt {41}/5}\bigg|+c$
=>$ \large\frac{1}{\sqrt {41}}$$ \log \bigg| \large\frac{t+4/5 - \sqrt {41}/5}{t- 4/5 +\sqrt {41}/5}\bigg|+c$
Hence c is the correct answer.
answered Jan 2, 2014 by meena.p
 
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