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Evaluate the indefinite integral:
\(\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}\)
Trigonometric substitution:
\(\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}\)
then,
\(\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}\)
So the integral \(\mathsf{(ii)}\) becomes
\(\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}\)
Integrate \(\mathsf{(ii)}\) by parts:
\(\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}\)
\(\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}\)
Substitute back for the variable x, and you get
\(\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}\)
I hope this helps. =)
Tags: integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus
