Mr. Franklin asked each of his 30 students to
2 answers:
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Evaluate the indefinite integral:
Trigonometric substitution:
then,
![\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcl%7D%20%5Cmathsf%7Bx%3Dsin%5C%2C%5Ctheta%7D%26%5Cquad%5CRightarrow%5Cquad%26%5Cmathsf%7Bdx%3Dcos%5C%2C%5Ctheta%5C%2Cd%5Ctheta%5Cqquad%5Ccheckmark%7D%5C%5C%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bx%5E2%3Dsin%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bx%5E2%3D1-cos%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bcos%5E2%5C%2C%5Ctheta%3D1-x%5E2%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Csqrt%7B1-x%5E2%7D%5Cqquad%5Ccheckmark%7D%5C%5C%5C%5C%5C%5C%20%26%26%5Ctextsf%7Bbecause%20%7D%5Cmathsf%7Bcos%5C%2C%5Ctheta%7D%5Ctextsf%7B%20is%20positive%20for%20%7D%5Cmathsf%7B%5Ctheta%5Cin%20%5Cleft%5B%5Cdfrac%7B%5Cpi%7D%7B2%7D%2C%5C%2C%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D.%7D%20%5Cend%7Barray%7D)
So the integral
becomes
Integrate by parts:
Substitute back for the variable x, and you get
I hope this helps. =)
Tags: <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>
_______________
Evaluate the indefinite integral:
Trigonometric substitution:
then,
So the integral
Integrate
Substitute back for the variable x, and you get
I hope this helps. =)
Tags: <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>