Question

integrate sin x / root 4 cos x^2 -1​

Answer 1

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To evaluate

$∫  \frac{sinx}{ \sqrt{4 {cos}^{2}x - 1 } } dx$

Formula used :-

$∫  \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log(x + \sqrt{ {x}^{2} - {a}^{2} } ) + c$

Solution:-

$∫  \frac{sinx}{ \sqrt{4 {cos}^{2}x - 1 } } \\ \small\bold\red{put \: cosx \: = y} \\ \small\bold\red{ - sinx = \frac{dy}{dx} } \\ \small\bold\red{ = > \: sinx \: dx \: = - dy}$

$so \: integral \: reduces \: to \\   = ∫  \frac{ - dy}{ \sqrt{4 {y}^{2} - 1 } } \\ = - \frac{1}{2}  ∫  \frac{dy}{ \sqrt{ {y}^{2} - \frac{1}{4} } } \\ = - \frac{1}{2}  ∫  \frac{dy}{ \sqrt{ {y}^{2} - {( \frac{1}{2}) }^{2} } } \\ = - \frac{1}{2} log |y + \sqrt{ ({y}^{2} - \frac{1}{4}) } | + c \\ = - \frac{1}{2} log |y + \sqrt{( \frac{ {4y}^{2} - 1}{4} )} | + c \\ = - \frac{1}{2} log | \frac{2y + \sqrt{ {4y}^{2} - 1} }{2} | + c \\ = - \frac{1}{2} ( log |2y + \sqrt{( {4y}^{2} - 1)} | - log(2) ) + c \\ = - \frac{1}{2} log(y + \sqrt{ {4y}^{2} - 1} ) - \frac{1}{2} log(2) + c \\ = - \frac{1}{2} log(cosx + \sqrt{ {4cos}^{2}x - 1 } ) + d$

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