# How can I solve this trigonometric integral step by step?

I have already posted a question similar to this one so I'm sorry if you're seeing this again. However, I'd like to whether I could get help solving the below definite integral step by step because I do not understand how to get to the answer which my GDC calculator gives me. The problem involves a trigonometric integral with a sine to the second degree. The problem goes as follows: $$\pi\int_0^{109.620851326}(8.89412\sin(0.047169x+0.105306)+21.8786)^2~dx$$ My attempts at the solution to both the indefinite and definite integral involved the expression for the reduction of $$\sin^2$$ which is the correct approach. I reduced sine in the following manner: $$\sin^2\theta=\frac{1-\cos{2\theta}}{2}$$ The answer to the definite integral also seems to be 15086.383263155 so the goal is to reach that answer. Whatever I seem to do I always get around 642.78185152953. I know that it will be difficult to explain every step and for that I apologise, I just haven't been able to find help to do this anywhere else. I really appreciate any help! :) Thank you.

P.S Sorry for the poor MathJax, I'm new to the system .

• That many decimal places is completely unnecessary. – Don Thousand Oct 28 '19 at 17:06
• I agree with you. My plan was to eliminate the decimal places after I explain everything I had to in order to see if the answer was right. – S_Orbital Oct 28 '19 at 17:20

\begin{align} k\int_a^b \frac{1-\cos2\theta}2\ d\theta&=k\int_a^b\frac12-\frac12\cos2\theta\ d\theta\\ &=k\bigg(\frac12\theta-\frac12\cdot\frac12\sin2\theta\bigg)\bigg]_a^b\\ &=k\bigg(\big(\frac12b-\frac14\sin2b\big)-\big(\frac12a-\frac14\sin2a\big)\bigg) \end{align}

What is the expression to which you reduced?

Edit: the expression that you want to reduce is in the form $$(\alpha\sin(\beta x+\gamma)+\delta)^2$$

The manipulation should look like \begin{align} (\alpha\sin(\beta x+\gamma)+\delta)^2&=\alpha^2\sin^2(\beta x+\gamma)+2\alpha\delta\sin(\beta x+\gamma)+\delta^2\\ &=\alpha^2\bigg(\frac{1-\cos2(\beta x+\gamma)}{2}\bigg)+2\alpha\delta\sin(\beta x+\gamma)+\delta^2\\ &=\frac{\alpha^2}{2}+\delta^2-\frac{\alpha^2}{2}\cos2(\beta x+\gamma)+2\alpha\delta\sin(\beta x+\gamma) \end{align}

Then, letting $$I$$ be the definite integral we need above, we have \begin{align} I&=\pi\int_a^b \frac{\alpha^2}{2}+\delta^2-\frac{\alpha^2}{2}\cos2(\beta x+\gamma)+2\alpha\delta\sin(\beta x+\gamma)\ dx\\ &=\pi\bigg((\frac{\alpha^2}{2}+\delta^2)x-\frac{\alpha^2}{4\beta}\sin 2(\beta x+\gamma)-\frac{2\alpha\delta}{\beta}\cos(\beta x+\gamma)\bigg)\bigg]_a^b \end{align}

Make the appropriate substitutions, and you have your final answer.

• The end product of me integrating the reduced version went as follows: $$\frac{0.0047169x+0.105306}{2}-\frac{\sin(2(0.0047169x+0.105306))}{4}$$ and the final product, with the above expression being multiplied by the constants outside of the integral at the beginning was as follows: $$5.861145673731x+13.085157758653-\frac{\sin(2(0.0047169x+0.105306))}{4}+C$$ – S_Orbital Oct 28 '19 at 17:24
• @S_Orbital As pointed out in a different comment that so many decimal places are unnecessary, I have made substitutions to illustrate to what value the integral should evaluate. – Andrew Chin Oct 28 '19 at 17:36