Question

integrate dx/1+3cos x =​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\tt \ \: :  ⟼ (1) \: cosx \: = \dfrac{1 - {tan}^{2}\dfrac{x}{2}}{1 + {tan}^{2}\dfrac{x}{2}}$

$\tt \ \: :  ⟼ (2) \: \: \int \dfrac{dx}{ {a}^{2} - {x}^{2} } = \dfrac{1}{2a} log(\dfrac{a + x}{a - x} ) + c$

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$\large\underline\purple{\bold{Solution :- }}$

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$\tt \ \: :  ⟼ \: Let \: I \: = \int \dfrac{dx}{1 + 3cosx}$

$\tt \ \: :  ⟼ \: I = \int \dfrac{dx}{1 + 3 \bigg(\dfrac{1 - {tan}^{2}\dfrac{x}{2} }{1 + {tan}^{2}\dfrac{x}{2}} \bigg)}$

☆ Now, taking LCM, we get

$\tt \ \: :  ⟼ \:I = \int \dfrac{dx}{ \bigg( \dfrac{1 + {tan}^{2}\dfrac{x}{2} + 3 - 3 {tan}^{2}\dfrac{x}{2}}{1 + {tan}^{2}\dfrac{x}{2}} \bigg)}$

$\tt \ \: :  ⟼ \: I = \int \dfrac{1 + {tan}^{2}\dfrac{x}{2}}{4 - 2{tan}^{2}\dfrac{x}{2}} \: \: dx$

$\tt \ \: :  ⟼ \: I = \dfrac{1}{2} \int \dfrac{ {sec}^{2}\dfrac{x}{2} \: \: \: dx}{2 - {tan}^{2}\dfrac{x}{2}}$

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$\tt \ \: :  ⟼ Now, \: put \: tan\dfrac{x}{2} = y$

$\tt \:  ⟼ differentiate \: w.r.t. \: x \: we \: get$

$\tt \:  ⟼ {sec}^{2} \dfrac{x}{2} \times \dfrac{1}{2} = \dfrac{dy}{dx}$

$\tt\implies \: {sec}^{2} \dfrac{x}{2} dx = 2dy$

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$\tt \ \: :  ⟼ I \: = \dfrac{1}{2} \int\dfrac{2 \: dy}{2 - {y}^{2} }$

$\tt \ \: :  ⟼ \: I = \int \dfrac{dy}{ {( \sqrt{2} )}^{2} - {y}^{2} }$

$\tt \ \: :  ⟼ I = \dfrac{1}{2 \sqrt{2} } log \bigg(\dfrac{ \sqrt{2} + y}{ \sqrt{2} - y} \bigg) + c$

$\tt \ \: :  ⟼ I = \dfrac{1}{2 \sqrt{2} } log \bigg(\dfrac{ \sqrt{2} + tan\dfrac{x}{2} }{ \sqrt{2} - tan\dfrac{x}{2} } \bigg) + c$

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