Evaluate the following intregals: 1 4 -1 dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{1}{4\cos\text{x}-1}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{4\cos\text{x}-1}\ \text{dx}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1-\tan^2\frac{\text{x}}{2}}$
$\Rightarrow\text{I}=\int\frac{1}{4\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)-1}\ \text{dx}$
$=\int\frac{1}{\frac{4\Big(1-\tan^2\frac{\text{x}}{2}\Big)-\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)\text{dx}}{4-4\tan^2\big(\frac{\text{x}}{2}\big)-1-\tan^2\big(\frac{\text{x}}{2}\big)}$
$=\int\frac{\sec^2\big(\frac{\text{x}}{2}\big)\text{dx}}{3-5\tan^2\big(\frac{\text{x}}{2}\big)}$
Let $\tan\Big(\frac{\text{x}}{2}\Big)=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$
$\therefore\ \text{I}=2\int\frac{\text{x}}{3-5\text{t}^2}$
$=\frac{2}{5}\int\frac{\text{dt}}{\frac{3}{5}-\text{t}^2}$
$=\frac{2}{5}\int\frac{\text{dt}}{\Big(\frac{\sqrt{3}}{\sqrt{5}}\Big)^2-\text{t}^2}$
$=\frac{2}{5}\times\frac{\sqrt{5}}{2\sqrt{3 }}\ln\begin{vmatrix}\frac{\frac{\sqrt{3}}{\sqrt{5}}+\text{t}}{\frac{\sqrt{3}}{\sqrt{5}}-\text{t}}\end{vmatrix}+\text{C}$
$=\frac{1}{\sqrt{15}}\ln\bigg|\frac{\sqrt{3}+\sqrt{5}\text{t}}{\sqrt{3}-\sqrt{5}\text{t}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{15}}\ln\begin{vmatrix}\frac{\sqrt{3}+\sqrt{5}\tan\big(\frac{\text{x}}{2}\big)}{\sqrt{3}-\sqrt{5}\tan\big(\frac{\text{x}}{2}\big)}\end{vmatrix}+\text{C}$

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