Problem 12

Question

Use integration tables to find the integral. $$ \int \frac{\theta^{2}}{1-\sin \theta^{3}} d \theta $$

Step-by-Step Solution

Verified
Answer
\((1/3)*\tan^{-1}(\sin(\theta^{3})) + C\)
1Step 1: Substitution
Use the property of substitution. A good choice here is to set \(u = \sin(\theta^{3})\). Differentiating both sides gives \(du = 3\theta^{2}\cos(\theta^{3})d\theta\).
2Step 2: Rearrange Equation
Rearrange this equation to express \(\theta^{2}d\theta\), we get \(\theta^{2}d\theta = (1/3)du/\cos(\theta^{3}))\). Substituting this into the original equation leads to a simplified integral \(\int (1/3)\frac{1}{1-u} du/\cos(\theta^{3})\).
3Step 3: Given condition by exercise
The original problem does not give information about \(\cos(\theta^{3})\). What should be noted is that \(cos^{2}(x)=1−sin^{2}(x)\), or \(\cos(\theta^{3}) = \sqrt{1-u^{2}}\). This leads to \(\int (1/3)\frac{1}{1-u} du/\sqrt{1-u^{2}}\).
4Step 4: Simplification and look up an integral table
After obtaining the simplified form of the integral which is necessary to compute, check an integral table. The integral of \(\frac{1}{a^2-u^2}\), with \(a = 1\) is a standard table entry leading to \(1/a \tan^{-1}(u/a) + C\), here \(C\) denotes the constant of integration. The integral of \(\frac{1}{1-u^2}\) is the equivalent of the aforementioned standard form. Apply it and we get \((1/3)*(1/1)\tan^{-1}(u/1) + C\).
5Step 5: Revert to the original variable
The final step is to revert to the original variable \(u\). Substitute \(u = \sin(\theta^{3})\) back in and the final answer is \((1/3)*\tan^{-1}(\sin(\theta^{3})) + C\).