Evaluate: `int1/(5+4cosx)dx`

Explanation

Let` I=int frac{1}{13+3 cos x+4 sin x} d x`
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Putting cos x=frac{1-tan ^{2} frac{x}{2}}{1+tan ^{2} frac{x}{2}} and sin x=frac{2 tan (frac{x}{2})}{1+tan ^{2}(frac{x}{2})}`
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I=int frac{1}{13+3(frac{1-tan ^{2} frac{z}{4}}{1+tan ^{2} frac{2}{2}})+4 times 2 frac{tan (frac{z}{3})}{1+tan ^{2}(frac{2}{2})}} d x`
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=int frac{(1+tan ^{2} frac{x}{2})}{13(1+tan ^{2} frac{x}{2})+3-3 tan ^{2} frac{x}{2}+8 tan (frac{x}{2})} d x`
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=int frac{sec ^{2} frac{x}{2}}{13 tan ^{2} frac{x}{2}-3 tan ^{2} frac{x}{2}+16+8 tan (frac{x}{2})} d x`
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=int frac{sec ^{2} frac{x}{2}}{10 tan ^{2}(frac{x}{2})+8 tan }`
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Let tan (frac{x}{2})=t`
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Rightarrow frac{1}{2} sec ^{2}(frac{x}{2}) d x=d t`
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