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Find the indefinite integral of \(\displaystyle \int \frac{12 \cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx\).
Making the u substitution \(\displaystyle u = \sin{\left(x \right)}\) gives \(\displaystyle du = \left(\cos{\left(x \right)}\right)dx\). Solving for \(\displaystyle dx\) gives \(\displaystyle dx = \frac{1}{\cos{\left(x \right)}}du\). Substituting in the values of \(\displaystyle u\) and \(\displaystyle du\) gives \(\displaystyle \int \left(\frac{12 \cos{\left(x \right)}}{u^{3}}\right)\left(\frac{1}{\cos{\left(x \right)}}du\right)\). Simplifying gives the integral \(\displaystyle \int \frac{12}{u^{3}} du\). Integrating gives \(\displaystyle \int \frac{12}{u^{3}} du = - \frac{6}{u^{2}}+C\). Substituting \(\displaystyle u\) back in gives the solution \(\displaystyle - \frac{6}{\sin^{2}{\left(x \right)}}+C\).
\begin{question}Find the indefinite integral of $\int \frac{12 \cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx$. \soln{9cm}{Making the u substitution $u = \sin{\left(x \right)}$ gives $du = \left(\cos{\left(x \right)}\right)dx$. Solving for $dx$ gives $dx = \frac{1}{\cos{\left(x \right)}}du$. Substituting in the values of $u$ and $du$ gives $\int \left(\frac{12 \cos{\left(x \right)}}{u^{3}}\right)\left(\frac{1}{\cos{\left(x \right)}}du\right)$. Simplifying gives the integral $\int \frac{12}{u^{3}} du$. Integrating gives $\int \frac{12}{u^{3}} du = - \frac{6}{u^{2}}+C$. Substituting $u$ back in gives the solution $- \frac{6}{\sin^{2}{\left(x \right)}}+C$. } \end{question}
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<p> <p>Find the indefinite integral of <img class="equation_image" title=" \displaystyle \int \frac{12 \cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%20%5Cfrac%7B12%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Csin%5E%7B3%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%5C%2C%20dx%20" alt="LaTeX: \displaystyle \int \frac{12 \cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx " data-equation-content=" \displaystyle \int \frac{12 \cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx " /> . </p> </p>
<p> <p>Making the u substitution <img class="equation_image" title=" \displaystyle u = \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle u = \sin{\left(x \right)} " data-equation-content=" \displaystyle u = \sin{\left(x \right)} " /> gives <img class="equation_image" title=" \displaystyle du = \left(\cos{\left(x \right)}\right)dx " src="/equation_images/%20%5Cdisplaystyle%20du%20%3D%20%5Cleft%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%5Cright%29dx%20" alt="LaTeX: \displaystyle du = \left(\cos{\left(x \right)}\right)dx " data-equation-content=" \displaystyle du = \left(\cos{\left(x \right)}\right)dx " /> . Solving for <img class="equation_image" title=" \displaystyle dx " src="/equation_images/%20%5Cdisplaystyle%20dx%20" alt="LaTeX: \displaystyle dx " data-equation-content=" \displaystyle dx " /> gives <img class="equation_image" title=" \displaystyle dx = \frac{1}{\cos{\left(x \right)}}du " src="/equation_images/%20%5Cdisplaystyle%20dx%20%3D%20%5Cfrac%7B1%7D%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7Ddu%20" alt="LaTeX: \displaystyle dx = \frac{1}{\cos{\left(x \right)}}du " data-equation-content=" \displaystyle dx = \frac{1}{\cos{\left(x \right)}}du " /> . Substituting in the values of <img class="equation_image" title=" \displaystyle u " src="/equation_images/%20%5Cdisplaystyle%20u%20" alt="LaTeX: \displaystyle u " data-equation-content=" \displaystyle u " /> and <img class="equation_image" title=" \displaystyle du " src="/equation_images/%20%5Cdisplaystyle%20du%20" alt="LaTeX: \displaystyle du " data-equation-content=" \displaystyle du " /> gives <img class="equation_image" title=" \displaystyle \int \left(\frac{12 \cos{\left(x \right)}}{u^{3}}\right)\left(\frac{1}{\cos{\left(x \right)}}du\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%20%5Cleft%28%5Cfrac%7B12%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%7Bu%5E%7B3%7D%7D%5Cright%29%5Cleft%28%5Cfrac%7B1%7D%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7Ddu%5Cright%29%20" alt="LaTeX: \displaystyle \int \left(\frac{12 \cos{\left(x \right)}}{u^{3}}\right)\left(\frac{1}{\cos{\left(x \right)}}du\right) " data-equation-content=" \displaystyle \int \left(\frac{12 \cos{\left(x \right)}}{u^{3}}\right)\left(\frac{1}{\cos{\left(x \right)}}du\right) " /> . Simplifying gives the integral <img class="equation_image" title=" \displaystyle \int \frac{12}{u^{3}} du " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%20%5Cfrac%7B12%7D%7Bu%5E%7B3%7D%7D%20du%20" alt="LaTeX: \displaystyle \int \frac{12}{u^{3}} du " data-equation-content=" \displaystyle \int \frac{12}{u^{3}} du " /> . Integrating gives <img class="equation_image" title=" \displaystyle \int \frac{12}{u^{3}} du = - \frac{6}{u^{2}}+C " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%20%5Cfrac%7B12%7D%7Bu%5E%7B3%7D%7D%20du%20%3D%20-%20%5Cfrac%7B6%7D%7Bu%5E%7B2%7D%7D%2BC%20" alt="LaTeX: \displaystyle \int \frac{12}{u^{3}} du = - \frac{6}{u^{2}}+C " data-equation-content=" \displaystyle \int \frac{12}{u^{3}} du = - \frac{6}{u^{2}}+C " /> . Substituting <img class="equation_image" title=" \displaystyle u " src="/equation_images/%20%5Cdisplaystyle%20u%20" alt="LaTeX: \displaystyle u " data-equation-content=" \displaystyle u " /> back in gives the solution <img class="equation_image" title=" \displaystyle - \frac{6}{\sin^{2}{\left(x \right)}}+C " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B6%7D%7B%5Csin%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%2BC%20" alt="LaTeX: \displaystyle - \frac{6}{\sin^{2}{\left(x \right)}}+C " data-equation-content=" \displaystyle - \frac{6}{\sin^{2}{\left(x \right)}}+C " /> . </p> </p>