Evaluate the following integrals: 5 ^3x+6 ^3x 2 ^2x ^2x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\text{dx}$

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Answer

$\int\bigg(\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\bigg(\frac{5\cos^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}+\frac{6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\Big(\frac{5}{2}\frac{\cos\text{x}}{\sin^2\text{x}}+3\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\frac{5}{2}\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}\Big)\text{dx}+3\int\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\text{dx}$
$=\frac{5}{2}\int(\text{cosec x}\cot\text{x})\text{dx}+3\int\sec\text{x}\tan\text{x dx}$
$=\frac{5}{2}(-\text{cosec x})+3\sec\text{x}+\text{C}$
$=-\frac{5}{2}\text{cosec x}+3\sec\text{x}+\text{C}$

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