Integration of dx/(3sinx+ 4cosx)?

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Integration of dx/(3sinx+ 4cosx)?

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Answer 1 · 2017-09-01T14:10:44

$\int \frac{d x}{3 sin x + 4 cos x} = - \frac{1}{5} ln | csc (x + α) + cot (x + α) | + c$ $intdx/(3sinx+4cosx)=-1/5ln|csc(x+alpha)+cot(x+alpha)|+c$ , where $α = {tan}^{- 1} (\frac{4}{3})$ $alpha=tan^(-1)(4/3)$

Explanation:

$3 sin x + 4 cos x$ $3sinx+4cosx$

= $5 (sin x \times \frac{3}{5} + cos x \times \frac{4}{5})$ $5(sinx xx 3/5+cosx xx 4/5)$

= $5 sin (x + α)$ $5sin(x+alpha)$ , where $tan α = \frac{4}{3}$ $tanalpha=4/3$ or $α = {tan}^{- 1} (\frac{4}{3})$ $alpha=tan^(-1)(4/3)$

Hence, $\int \frac{d x}{3 sin x + 4 cos x}$ $intdx/(3sinx+4cosx)$

= $\frac{1}{5} \int csc (x + α) d x$ $1/5intcsc(x+alpha)dx$

Now let $x + α = u$ $x+alpha=u$ then $d x = d u$ $dx=du$ and

$\frac{1}{5} \int csc (x + α) d x = \frac{1}{5} \int csc u d u$ $1/5intcsc(x+alpha)dx=1/5intcscudu$

= $\frac{1}{5} \int \frac{csc u (csc u + cot u)}{csc u + cot u} d u$ $1/5int(cscu(cscu+cotu))/(cscu+cotu)du$

= $- \frac{1}{5} \int \frac{- {csc}^{2} u - csc u cot u}{csc u + cot u} d u$ $-1/5int(-csc^2u-cscucotu)/(cscu+cotu)du$

if $csc u + cot u = v$ $cscu+cotu=v$ , then $d v = (- {csc}^{2} u - csc u cot u) d u$ $dv=(-csc^2u-cscucotu)du$

and our integral becomes

$- \frac{1}{5} \int \frac{d v}{v} = - \frac{1}{5} ln | v | + c$ $-1/5int(dv)/v=-1/5ln|v|+c$

= $- \frac{1}{5} ln | csc (x + α) + cot (x + α) | + c$ $-1/5ln|csc(x+alpha)+cot(x+alpha)|+c$ , where $α = {tan}^{- 1} (\frac{4}{3})$ $alpha=tan^(-1)(4/3)$

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Integration of dx/(3sinx+ 4cosx)?

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