I think ... the question is ill posed. Indefinite integrals don't have a "value". Sometimes you can't have a closed formula for some functions, yet for specific limit values you cand find out the exact integral value
–
Omm Yucatan
·
3 years ago

Assume \sin\theta =t
We have \cos\theta dø = dt,by difrentiat it
\cos\theta =\sqrt{t^{2}-1}
So new function is \frac{ \sin\theta }{\sqrt{t^{2}-1}
Solve this function by part ,may be we a solution
–
Prabhat Sharma
·
3 years ago

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@Prabhat Sharma
–
\[t=\sin\theta\]
\[\frac{\mbox{d}t}{\mbox{d}\theta}=\cos\theta\]
\[\mbox{d}\theta=\frac{\mbox{d}t}{\sqrt{1-t^2}}\]
\[\begin{array}{rcl}
I&=&\int\sin(\sin\theta)\mbox{ d}\theta\\
&=&\int\frac{\sin t\mbox{ d}t}{\sqrt{1-t^2}}
\end{array}\]
–
Kenny Lau
·
3 years ago

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TopNewestI think ... the question is ill posed. Indefinite integrals don't have a "value". Sometimes you can't have a closed formula for some functions, yet for specific limit values you cand find out the exact integral value – Omm Yucatan · 3 years ago

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– Kenny Lau · 3 years ago

Thanks for your reminder. I've updated the question.Log in to reply

Assume \sin\theta =t We have \cos\theta dø = dt,by difrentiat it \cos\theta =\sqrt{t^{2}-1} So new function is \frac{ \sin\theta }{\sqrt{t^{2}-1} Solve this function by part ,may be we a solution – Prabhat Sharma · 3 years ago

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– Kenny Lau · 3 years ago

\[t=\sin\theta\] \[\frac{\mbox{d}t}{\mbox{d}\theta}=\cos\theta\] \[\mbox{d}\theta=\frac{\mbox{d}t}{\sqrt{1-t^2}}\] \[\begin{array}{rcl} I&=&\int\sin(\sin\theta)\mbox{ d}\theta\\ &=&\int\frac{\sin t\mbox{ d}t}{\sqrt{1-t^2}} \end{array}\]Log in to reply