Question

Evaluate the integral\(\)\(\int {\frac{{{\mathop{\rm Cos}\nolimits} x}}{{{{{\mathop{\rm Sin}\nolimits} }^2}x - 9}}} .dx\)

Short answer

\(\int {\frac{{{\mathop{\rm Cos}\nolimits} x}}{{{{{\mathop{\rm Sin}\nolimits} }^2}x - 9}}} .dx = \frac{1}{6}\log |\frac{{\sin x - 3}}{{\sin x + 3}}| + c\)

Step-by-step solution

Step 1- (Find the integral)

\(\frac{{{\mathop{\rm Cos}\nolimits} x}}{{{{\sin }^2}x - 9}}.dx\)

Let \({\mathop{\rm Sin}\nolimits} x = t\)

Differentiating both sides with respect to x

\(\)\(\begin{aligned}{l}{\mathop{\rm Cos}\nolimits} xdx = dt\\dx = \frac{{dt}}{{\cos x}}\end{aligned}\)

\(\begin{aligned}{l}\int {\frac{{{\mathop{\rm Cos}\nolimits} x}}{{{{{\mathop{\rm Sin}\nolimits} }^2}x - 9}}} .dx = \int {\frac{{{\mathop{\rm Cos}\nolimits} x}}{{{t^2} - {3^2}}}} .{\frac{{dt}}{{{\mathop{\rm Cos}\nolimits} x}}^{}}\\ = \int {\frac{1}{{{t^2} - {3^2}}}} .dt\end{aligned}\)

Using the formula,\(\)\(\begin{aligned}{l}\int {\frac{1}{{{x^2} - {a^2}}}} .dx = \frac{1}{{2a}}\log |\frac{{x - a}}{{x + a}}\frac{{ + c}}{{}}\\ = \frac{1}{{2*3}}\log |\frac{{t - 3}}{{t + 3}}| + c\\ = \frac{1}{6}\log |\frac{{{\mathop{\rm Sin}\nolimits} x - 3}}{{{\mathop{\rm Sin}\nolimits} x + 3}}| + c\end{aligned}\)

\(\)

Hence\(\int {\frac{{{\mathop{\rm Cos}\nolimits} x}}{{{{{\mathop{\rm Sin}\nolimits} }^2}x - 9}}} .dx = \frac{1}{6}\log |\frac{{{\mathop{\rm Sin}\nolimits} x - 3}}{{{\mathop{\rm Sin}\nolimits} x + 3}}| + c\)