what is integral of sin^(-1)
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Let I=∫sin−1xdxI=∫sin−1xdx
=∫1⋅sin−1xdx=∫1⋅sin−1xdx
by using integration by parts
I=sin−1x∫1dx−∫I=sin−1x∫1dx−∫(ddxsin−1x∫1dx)dx(ddxsin−1x∫1dx)dx
=xsin−1x−∫(x1−x2−−−−−√)dx=xsin−1x−∫(x1−x2)dx
=xsin−1x+∫−2x21−x2−−−−−√=xsin−1x+∫−2x21−x2
Put 1−x2=t1−x2=t
⟹−2xdx=dt⟹−2xdx=dt
∴I=xsin−1x+∫12t√∴I=xsin−1x+∫12t
=xsin−1x+t√+c=xsin−1x+t+c
⟹I=xsin−1x+1−x2−−−−−√+c⟹I=xsin−1x+1−x2+c
=∫1⋅sin−1xdx=∫1⋅sin−1xdx
by using integration by parts
I=sin−1x∫1dx−∫I=sin−1x∫1dx−∫(ddxsin−1x∫1dx)dx(ddxsin−1x∫1dx)dx
=xsin−1x−∫(x1−x2−−−−−√)dx=xsin−1x−∫(x1−x2)dx
=xsin−1x+∫−2x21−x2−−−−−√=xsin−1x+∫−2x21−x2
Put 1−x2=t1−x2=t
⟹−2xdx=dt⟹−2xdx=dt
∴I=xsin−1x+∫12t√∴I=xsin−1x+∫12t
=xsin−1x+t√+c=xsin−1x+t+c
⟹I=xsin−1x+1−x2−−−−−√+c⟹I=xsin−1x+1−x2+c
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d( costhete) it will help you for the solution
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d( costhete) it will help you for the solution
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